SQL lab

Department

Of

Computer Science & Engineering

Data Base Management Systems Lab Manual

BALAJI INSTITUTE OF TECHNOLOGY AND SCIENCE

Department of computer science & engineering

In-charge HOD Principal

Prepared by: Approved & Reviewed by:

Issued by: w.e.f Date:

Department of Computer Science and Engineering

BALAJI INSTITUTE OF TECHNOLOGY AND SCIENCE

Laknepally(V), Narsampet, Warangal

DEPARTMENT OF COMPUTER SCIENCE & ENGINEERING

Lab Manual for the Academic Year 2011-12

(In accordance with JNTU syllabus)

SUBJECT : DBMS LAB

STREAM : CSE

H.O.D

Balaji Institute of Technology and Science 2


INDEX

S. No Contents Page. no

1 Lab Objective4

2 Introduction About Lab5

3Standard Operating Procedure – SOP

7

3 Guidelines to Students 8

4Description about SQL, PL/SQL statements and D2K.

9

5List of Lab Exercises4.1 Syllabus Programs (JNTU)4.2 Additional Programs

10

6 Background Theory15

7 Solutions for Programs 21

8 PL-Sql Programs51

9 Additional Viva Question89

10 References95


Department of Computer Science and EngineeringLAB OBJECTIVE

Upon successful completion of this Lab the student will be able to:

Creating database objects

Modifying database objects

Manipulating the data

Retrieving the data from the database server

Performing database operations in a procedural manner using pl/sql

Performing database operations (create, update, modify, retrieve, etc.,)

using front-end tools like D2K.

Design and Develop applications like banking, reservation system, etc.,


Department of Computer Science and EngineeringINTRODUCTION ABOUT LAB

There are 66 systems ( Compaq Presario ) installed in this Lab. Their configurations are as follows:

Processor : AMD Athelon ™ 1.67 GHz

RAM : 256 MB

Hard Disk : 40 GB

Mouse : Optical Mouse

Network Interface card : Present

Software

All systems are configured in DUAL BOOT mode i.e., Students can boot from Windows XP or Linux as per their lab requirement. This is very useful for students because they are familiar with different Operating Systems so that they can execute their programs in different programming environments.

Each student has a separate login for database access

Oracle 9i client version is installed in all systems. On the server, account for each student has been created.

This is very useful because students can save their work (scenarios’, pl / sql programs, data related projects, etc) in their own accounts. Each student work is safe and secure from other students.

Latest Technologies like DOT NET and J2EE are installed in some systems. Before submitting their final project, they can start doing mini project from 2nd year onwards.

MASM (Macro Assembler) is installed in all the systems

Students can execute their assembly language programs using MASM. MASM is very useful students because when they execute their


Department of Computer Science and Engineeringprograms they can see contents of Processor Registers and how each instruction is being executed in the CPU.

Rational Rose Software is installed in some systems

Using this software, students can depict UML diagrams of their projects.

Softwares installed : C, C++, JDK1.5, MASM, OFFICE-XP, J2EE and DOT NET, Rational Rose.

Systems are provided for students in the 1:1 ratio.

Systems are assigned numbers and same system is allotted for students when they do the lab.



STANDARD OPERATING PROCEDURE – SOP

a) Explanation on today’s experiment by the concerned faculty using OHP/PPT covering the following aspects: 25 mins.

1) Name of the experiment/Aim

2) Software/Hardware required

3) Algorithm

4) Test Data

1) Valid data sets

2) Limiting value sets

3) Invalid data sets

b) Writing of source program by the students 25 min.

c) Compiling and execution of the program

100 mins.

Writing of the experiment in the Observation Book :

The students will write the today’s experiment in the Observation book as per the following format:

a) Name of the experiment/Aim

b) Software/Hardware required

c) Algorithm

d) Source Program

e) Test Data

a. Valid data sets

b. Limiting value sets

c. Invalid data sets

f) Results for different data sets

g) Viva-Voc Questions and Answers

h) Errors observed (if any) during compilation/execution

i) Signature of the Faculty


Department of Computer Science and EngineeringGuidelines to Students

Equipment in the lab for the use of student community. Students need to maintain a proper decorum in the computer lab. Students must use the equipment with care. Any damage is caused is punishable.

Students are required to carry their observation / programs book with completed exercises while entering the lab.

Students are supposed to occupy the machines allotted to them and are not supposed to talk or make noise in the lab. The allocation is put up on the lab notice board.

Lab can be used in free time / lunch hours by the students who need to use the systems should take prior permission from the lab in-charge.

Lab records need to be submitted on or before date of submission.

Students are not supposed to use floppy disks


Department of Computer Science and EngineeringHow to Write and execute sql, pl/sql commands/programs:

1). Open your oracle application by the following navigationStart->all programs->oracle orahome.->application

development->sql.

2). You will be asked for user name, pass word and host stringYou have to enter user name, pass word and host string as given by the administrator. It will be different from one user to another user.

3). Upon successful login you will get SQL prompt (SQL>). In two ways you can write your programs: a). directly at SQL prompt b). or in sql editor.

If you type your programs at sql prompt then screen will look like follow:SQL> SELECT ename,empno,

2 sal from 3 emp;

where 2 and 3 are the line numbers and rest is the command /program……

to execute above program/command you have to press ‘/’ then enter.

Here editing the program is somewhat difficult; if you want to edit the previous command then you have to open sql editor (by default it displays the sql buffer contents). By giving ‘ed’ at sql prompt.(this is what I mentioned as a second method to type/enter the program).in the sql editor you can do all the formatting/editing/file operations directly by selecting menu options provided by it.

To execute the program which saved; do the followingSQL> @ programname.sqlOrSQL> Run programname.sql Then press ‘\’ key and enter.

This how we can write, edit and execute the sql command and programs.Always you have to save your programs in your own logins.


Department of Computer Science and EngineeringList of Lab Exercises

Syllabus Programs (JNTU)

S. No Name of the program1 Database Schema for a customer-sale scenario

Customer(Cust id : integer, cust_name: string)Item(item_id: integer, item_name: string, price: integer)Sale(bill_no: integer, bill_data: date, cust_id: integer, item_id: integer, qty_sold: integer)

For the above schema, perform the following—a) Create the tables with the appropriate integrity constraints

b) Insert around 10 records in each of the tables

c) List all the bills for the current date with the customer names and item numbers

d) List the total Bill details with the quantity sold, price of the item and the final amount

e) List the details of the customer who have bought a product which has a price>200

f) Give a count of how many products have been bought by each customer

g) Give a list of products bought by a customer having cust_id as 5

h) List the item details which are sold as of today

i) Create a view which lists out the bill_no, bill_date, cust_id, item_id, price, qty_sold, amountCreate a view which lists the daily sales date wise for the last one week

2 Database Schema for a Student Library scenario

Student(Stud_no : integer, Stud_name: string)Membership(Mem_no: integer, Stud_no: integer)Book(book_no: integer, book_name:string, author: string)Iss_rec(iss_no:integer, iss_date: date, Mem_no: integer, book_no: integer)

For the above schema, perform the following—a) Create the tables with the appropriate integrity constraintsb) Insert around 10 records in each of the tablesc) List all the student names with their membership numbers


Department of Computer Science and Engineeringd) List all the issues for the current date with student and Book namese) List the details of students who borrowed book whose author is CJDATEf) Give a count of how many books have been bought by each studentg) Give a list of books taken by student with stud_no as 5h) List the book details which are issued as of todayi) Create a view which lists out the iss_no, iss _date, stud_name, book namej) Create a view which lists the daily issues-date wise for the last one week

3 Database Schema for a Employee-pay scenario

employee(emp_id : integer , emp_name: string)department(dept_id: integer, dept_name:string)paydetails(emp_id : integer, dept_id: integer, basic: integer, deductions: integer, additions: integer, DOJ: date)payroll(emp_id : integer, pay_date: date)

For the above schema, perform the following—a) Create the tables with the appropriate integrity constraintsb) Insert around 10 records in each of the tablesc) List the employee details department wised) List all the employee names who joined after particular date e) List the details of employees whose basic salary is between 10,000 and 20,000f) Give a count of how many employees are working in each departmentg) Give a names of the employees whose netsalary>10,000 h) List the details for an employee_id=5i) Create a view which lists out the emp_name, department, basic, dedeuctions, netsalaryj) Create a view which lists the emp_name and his netsalary

4 Database Schema for a Video Library scenario

Customer(cust_no: integer,cust_name: string)Membership(Mem_no: integer, cust_no: integer)Cassette(cass_no:integer, cass_name:string, Language: String) Iss_rec(iss_no: integer, iss_date: date, mem_no: integer, cass_no: integer)For the above schema, perform the following—a) Create the tables with the appropriate integrity constraintsb) Insert around 10 records in each of the tables


Department of Computer Science and Engineeringc) List all the customer names with their membership numbersd) List all the issues for the current date with the customer names and cassette namese) List the details of the customer who has borrowed the cassette whose title is “ The Legend”f) Give a count of how many cassettes have been borrowed by each customerg) Give a list of book which has been taken by the student with mem_no as 5h) List the cassettes issues for todayi) Create a view which lists outs the iss_no, iss_date, cust_name, cass_namej) Create a view which lists issues-date wise for the last one week

5 Database Schema for a student-Lab scenario

Student(stud_no: integer, stud_name: string, class: string)Class(class: string, descrip: string)Lab(mach_no: integer, Lab_no: integer, description: String)Allotment(Stud_no: Integer, mach_no: integer, dayof week: string)For the above schema, perform the following—a) Create the tables with the appropriate integrity constraintsb) Insert around 10 records in each of the tablesc) List all the machine allotments with the student names, lab and machine numbersd) List the total number of lab allotments day wisee) Give a count of how many machines have been allocated to the ‘CSIT’ classf) Give a machine allotment etails of the stud_no 5 with his personal and class detailsg) Count for how many machines have been allocatedin Lab_no 1 for the day of the week as “Monday”h) How many students class wise have allocated machines in the labsi) Create a view which lists out the stud_no, stud_name, mach_no, lab_no, dayofweekj) Create a view which lists the machine allotment details for “Thursday”.

6 Write a program to find largest number from the given three numbers.

7 Simple programs using loop, while and for iterative control statement.

8 Write a program to check whether the given number is Armstrong or not

9 Write a program to generate all prime numbers below 100.


Department of Computer Science and Engineering10 Write a program to demonstrate the GOTO statement.11 Write a program to demonstrate %type and %rowtype attributes12 Write a program to demonstrate predefined exceptions13 Write a program to demonstrate user defined exceptions14 Create a cursor, which displays all employee numbers and

names from the EMP table. 15 Create a cursor, which update the salaries of all employees as

per the given data.16 Create a cursor, which displays names of employees having

salary > 50000.17 Create a procedure to find reverse of a given number18 Create a procedure to update the salaries of all employees as

per the given data19 Create a procedure to demonstrate IN, OUT and INOUT

parameters20 Create a function to check whether given string is palindrome or

not.21 Create a function to find sum of salaries of all employees

working in depart number 10.22 Create a trigger before/after update on employee table for each

row/statement.23 Create a trigger before/after delete on employee table for each

row/statement.24 Create a trigger before/after insert on employee table for each

row/statement.25 Create a Form to display employee details using SQL26 Create a Report to generate all employee annual salaries….

Additional Programs

S. No Name of the Program

1 Create a form using Forms 6i to display Employee table data.

2 Create a Master/details relationship form which perform Add New, Search, Delete, Save and Update on the records

3 Generate a report to calculate employee’s salaries department wise from employee table.

4 Create a Report to generate the details of employee table including sum and average salaries department wise.


Department of Computer Science and EngineeringBackground Theory

Oracle workgroup or server is the largest selling RDBMS product.it is estimated that the combined sales of both these oracle database product account for aroud 80% of the RDBMSsystems sold worldwide.These products are constantly undergoing change and evolving. The natural language of this RDBMS product is ANSI SQL,PL/SQL a superset of ANSI SQL.oracle 8i and 9i also under stand SQLJ.

Oracle corp has also incorporated a full-fledged java virtual machine into its database engine.since both executable share the same memory space the JVM can communicate With the database engine with ease and has direct access to oracle tables and their data.

SQL is structure query language.SQL contains different data types those are1. char(size)2. varchar2(size)3. date4. number(p,s)5. long6. raw/long raw

Different types of commands in SQL:

A).DDL commands: - To create a database objectsB).DML commands: - To manipulate data of a database objectsC).DQL command: - To retrieve the data from a database.D).DCL/DTL commands: - To control the data of a database…

DDL commands:

1. The Create Table Command: - it defines each column of the table uniquely. Each column has minimum of three attributes, a name , data type and size.

Syntax:Create table <table name> (<col1> <datatype>(<size>),<col2> <datatype><size>));

Ex: create table emp(empno number(4) primary key, ename char(10));

2. Modifying the structure of tables.a)add new columns

Syntax:Alter table <tablename> add(<new col><datatype(size),<new col>datatype(size));



Ex:alter table emp add(sal number(7,2));

3. Dropping a column from a table.

Syntax:Alter table <tablename> drop column <col>;

Ex:alter table emp drop column sal;

4. Modifying existing columns.

Syntax:Alter table <tablename> modify(<col><newdatatype>(<newsize>));

Ex:alter table emp modify(ename varchar2(15));

5. Renaming the tables

Syntax:Rename <oldtable> to <new table>;

Ex:rename emp to emp1;

6. truncating the tables.

Syntax:Truncate table <tablename>;

Ex:trunc table emp1;

7. Destroying tables.

Syntax:Drop table <tablename>;

Ex:drop table emp;

DML commands:


Department of Computer Science and Engineering8. Inserting Data into Tables: - once a table is created the most natural thing to do is load this table with data to be manipulated later.

Syntax:insert into <tablename> (<col1>,<col2>) values(<exp>,<exp>);

9. Delete operations.

a) remove all rowsSyntax: delete from <tablename>;

b) removal of a specified row/sSyntax: delete from <tablename> where <condition>;

10. Updating the contents of a table.

a) updating all rowsSyntax:Update <tablename> set <col>=<exp>,<col>=<exp>;

b) updating seleted records.Syntax:Update <tablename> set <col>=<exp>,<col>=<exp>

where <condition>;

11. Types of data constrains.a) not null constraint at column level.Syntax:<col><datatype>(size)not null

b) unique constraintSyntax:Unique constraint at column level.<col><datatype>(size)unique;

c) unique constraint at table level:Syntax:Create table tablename(col=format,col=format,unique(<col1>,<col2>);

d) primary key constraint at column levelSyntax:<col><datatype>(size)primary key;

e) primary key constraint at table level.Syntax:Create table tablename(col=format,col=format primary key(col1>,<col2>);



f) foreign key constraint at column level.Syntax:<col><datatype>(size>) references <tablename>[<col>];

g) foreign key constraint at table level Syntax:

foreign key(<col>[,<col>])references <tablename>[(<col>,<col>)

h) check constraintcheck constraint constraint at column level.Syntax: <col><datatype>(size) check(<logical expression>)

i) check constraint constraint at table level.Syntax: check(<logical expression>)


Department of Computer Science and EngineeringDQL Commands:

12. Viewing data in the tables: - once data has been inserted into a table, the next most logical operation would be to view what has been inserted.

a) all rows and all columnsSyntax:

Select <col> to <col n> from tablename;

Select * from tablename;

13. Filtering table data: - while viewing data from a table, it is rare that all the data from table will be required each time. Hence, sql must give us a method of filtering out data that is not required data.

a) Selected columns and all rows:Syntax:select <col1>,<col2> from <tablename>;

b) selected rows and all columns:Syntax:select * from <tablename> where <condition>;

c) selected columns and selected rowsSyntax:select <col1>,<col2> from <tablename> where<condition>;

14. Sorting data in a table.Syntax:Select * from <tablename> order by <col1>,<col2> <[sortorder]>;

DCL commands:Oracle provides extensive feature in order to safeguard information stored in its tables from unauthoraised viewing and damage.The rights that allow the user of some or all oracle resources on the server are called privileges.

a) Grant privileges using the GRANT statement

The grant statement provides various types of access to database objects such as tables,views and sequences and so on.

Syntax:GRANT <object privileges>ON <objectname>TO<username>[WITH GRANT OPTION];


Department of Computer Science and Engineeringb) Reoke permissions using the REVOKE statement:

The REVOKE statement is used to deny the Grant given on an object.

Syntax:REVOKE<object privilege>ONFROM<user name>;



Solutions for programs (JNTU Syllabus)

1. Database Schema for a customer-sale scenario

Customer(Cust id : integer, cust_name: string)Item(item_id: integer, item_name: string, price: integer)Sale(bill_no: integer, bill_data: date, cust_id: integer, item_id: integer, qty_sold: integer)

For the above schema, perform the following—a) Create the tables with the appropriate integrity constraintsb) Insert around 10 records in each of the tablesc) List all the bills for the current date with the customer names and item

numbersd) List the total Bill details with the quantity sold, price of the item and the

final amounte) List the details of the customer who have bought a product which has a

price>200f) Give a count of how many products have been bought by each customerg) Give a list of products bought by a customer having cust_id as 5h) List the item details which are sold as of todayi) Create a view which lists out the bill_no, bill_date, cust_id, item_id, price,

qty_sold, amountj) Create a view which lists the daily sales date wise for the last one week


Department of Computer Science and EngineeringAim: Create the tables with the appropriate integrity constraints and Insert around 10 records in each of the tables

HW/SW requirements: Processor : AMD Athelon ™ 1.67 GHz

RAM : 256 MB

Hard Disk : 40 GB

Software : ORACLE

SQL> create table customer1 (cust_id number(5) primary key, cust_name

varchar2(15));Output: Table created.

SQL> desc customer1;

Output: Name Null? Type----------------------------------------- -------- ----------------CUST_ID NOT NULL NUMBER(5)CUST_NAME VARCHAR2(15)

Valid Test Data

b) SQL> insert into customer1 values(&custid,'&custname');SQL> select * from customer1;Output:CUST_ID CUST_NAME---------- ---------------

100 ramu 101 kamal 102 raju 103 raju sundaram 104 lawrence

SQL> create table item(item_id number(4) primary key, item_name varchar2(15),price number(6,2));SQL> dsec item

Output:Name Null? Type……………………………………………………………………………………………………Cust_id NOT NULL NUMBER(4)Item_name VARCHAR2(15)PRICE NUMBER(6,2)SQL>insert into item values(&item_id,’&item_name’,&price);


Department of Computer Science and Engineering SQL> select * from item;Output:ITEM_ID ITEM_NAME PRICE……………………………………………………………………………………..2334 geera 6.254532 corn soup 34.652124 lays chips 204531 setwet 99.992319 duracell 45.5

SQL>create table sale(bill_no number(5) primary key,bill_date date, cust_id number(5) references customer(cust_id), item_id number(4) references item(item_id),qty_sold number(4));

Out put: Table Created.

SQL>dsec saleOutput: Name Null? Type………………………………………………………………………………………..BILL_NO NOT NULL NUMBER(4)BILL_DATE DATECUST_ID NUMBER(5)ITEM_ID NUMBER(4)QTY_SOLD NUMBER(4)

SQL>insert into Sale values(&bill_no, ’&bill_date’, &cust_id, &item_id, &qty_sold);

SQL>select * from sale;Output:BILL_NO BILL_DATE CUST_ID ITEM_ID QTY_SOLD………………………………………………………………………………………………………...1450 04-JAN-06 100 2124 21451 04-JAN-06 101 2319 11452 04-JAN-06 103 4531 21453 04-JAN-06 102 2334 31454 04-JAN-06 104 4532 3

c) List all the bills for the current date with the customer names and item numbersSQL> select c.custname, i.itemid, s.billno from customer c, item I, sale s

where c.custid=s.custid ands.billdate=to_char(sysdate);

CUSTNAME ITEMID BILLNO------------- --------- ---------


Department of Computer Science and EngineeringJohn 5001 332

d) List the total Bill details with the quantity sold, price of the item and the final amountSQL> select i.price, s.qty,(i.price*s.qty) total from item I, sale s where

i.itemid=s.itemid;

PRICEQTY TOTAL------- ----- --------120 2 24020 3 605 2 1010 1 10350 4 1400

e) List the details of the customer who have bought a product which has a price>200

SQL> select c.custid, c.custname from customer c, sale s, item i where i.price>200 andc.custid=s.custid and i.itemid=s.itemid;

CUSTID CUSTNAME--------- --------------

4 duffy

f) Give a count of how many products have been bought by each customerSQL> select custid, count(itemid) from sale group by custid;

CUSTID COUNT(ITEMID)---------- ---------------------1 23 14 15 1

g) Give a list of products bought by a customer having cust_id as 5SQL> select i.itemname from item i, sale s where s.custid=5 and i.itemid-

s.itemid;

ITEMNAME--------------Pens

h) List the item details which are sold as of todaySQL> select i.itemid, i.itemname from item I, sale s where i.itemid=s.itemid


Department of Computer Science and Engineeringand s.billdate=to_char(sysdate);

ITEMID ITEMNAME--------- -------------

1234 pencil

i) Create a view which lists out the bill_no, bill_date, cust_id, item_id, price, qty_sold, amount

SQL>create view cust as (select s.billno, s.billdate, c.custid, i. iitemid, i.price, s.qty from customer c,sale s item I where c.custid=s.custid and i.iemid=s.itemid);

view created.

SQL>select * from cust;

BILLNO BILLDATE CUSTID ITEMID PRICE QTY……………………………………………………………………………………………3432 12-JAN-06 3 3244 120 24424 20-FEB-06 1 3456 20 3332 13-MAR-06 1 1234 5 22343 10-MAR 5 5001 10 11331 11-MAR-06 4 76776 350 4

j) Create a view which lists the daily sales date wise for the last one week

Viva-Voce:

Q1. What is SQL?Ans: Structured Query Language

2. What is database?A database is a logically coherent collection of data with some inherent

meaning, representing some aspect of real world and which is designed, built

and populated with data for a specific purpose.

3. What is DBMS?It is a collection of programs that enables user to create and maintain a

database. In other words it is general-purpose software that provides the users with the processes of defining, constructing and manipulating the database for various applications.

4. What is a Database system?The database and DBMS software together is called as Database system.


Department of Computer Science and Engineering5. Advantages of DBMS? Redundancy is controlled. Unauthorised access is restricted. Providing multiple user interfaces. Enforcing integrity constraints. Providing backup and recovery.

6. Disadvantage in File Processing System? Data redundancy & inconsistency. Difficult in accessing data. Data isolation. Data integrity. Concurrent access is not possible. Security Problems.


Department of Computer Science and Engineering2. Database Schema for a Student Library scenario

Student(Stud_no : integer, Stud_name: string)Membership(Mem_no: integer, Stud_no: integer)Book(book_no: integer, book_name:string, author: string)Iss_rec(iss_no:integer, iss_date: date, Mem_no: integer, book_no: integer)

For the above schema, perform the following—a) Create the tables with the appropriate integrity constraintsb) Insert around 10 records in each of the tablesc) List all the student names with their membership numbersd) List all the issues for the current date with student and Book namese) List the details of students who borrowed book whose author is CJDATEf) Give a count of how many books have been bought by each studentg) Give a list of books taken by student with stud_no as 5h) List the book details which are issued as of todayi) Create a view which lists out the iss_no, iss _date, stud_name, book namej) Create a view which lists the daily issues-date wise for the last one week


Department of Computer Science and EngineeringAIM: Create the tables with the appropriate integrity constraints Insert around 10 records in each of the tables


RAM : 256 MB

Hard Disk : 40 GB

Software : Oracle

SQL>create table student(stud_no number(5) primary key,stud_name varchar2(15));

SQL>desc student; Name Null? Type………………………………………………………………………………………..STUD_NO NOT NULL NUMBER(5)STUD_NAME VARCAHR2(15)Valid Test Data:

SQL>insert into student values(&stud_no,’&stud_name’);

SQL>select * from student;

STUD_NO STUD_NAME....................................................................

508 HARISH513 BALAJI518 RAKESH524 PAVAN534 JOYCE

SQL>create table membership(mem_no number(5) primary key,stud_no number(5) references student(stud)no));SQL>dsec membership;

Name Null? Type…………………………………………………………………………………………………….MEM_NO NOT NULL NUMBER(5)STUD_NO NUMBER(5)

SQL>insert into membership values(&mem_no,&stud_no);Enter value for mem_no:5440Enter value for stud_no:510old 1:insert into membership values(&mem_no,&stud_no)


Department of Computer Science and Engineeringnew 1:insert into membership values(5440,510)insert into membership values(5440,510)*Errors Observed:

ERROR at line 1:ORA-02291:integrity constraint(HARISH.SYS_C002724)violated-primary key not found

SQL>select * from membership; MEM_NO STUD_NO………………………………………………………………………..

5440 5135441 5085442 5185443 5345444 524

SQL>create table book(book_no number(5) primary key,book_name varchar2(20),author varchar2(2));

SQL>desc book;

Name Null? Type………………………………………………………………………………………..BOOK_NO NOT NULL NUMBER(5)BOOK_NAME VARCHAR2(20)AUTHOR VARCHAR2(20)

SQL>insert into book values(&book_no,’&book_name’,’&author’);SQL>select * from book;

BOOK_NO BOOK_NAME AUTHOR………………………………………………………………………………………………..9123 DBMS Rama Krishna2342 JAVA Robett wilkins4523 Fearless tales Alfred8723 my ambition Harish7821 Harry Potter JK Rowling

SQL>create table lss_rec(iss_no number primary key,iss_date date,mem_no number(5) references membership(mem_no),book_no number(5) references book(book_no));

SQL>desc iss_rec;Name Null? Type………………………………………………………………………………………………………ISS_NO NOT NULL NUMBERISS_DATE DATEMEM_NO NUMBER(5)


Department of Computer Science and EngineeringBOOK_NO NUMBER(5)

SQL>select * from iss_rec;ISS_NO ISS_DATE MEM_NO BOOK_NO…………………………………………………………………………………………………43 05-JAN-06 5443 452381 28-DEC-05 5441 872322 08-DEC-05 5440 782153 07-JAN-06 5442 912335 06-JAN-06 5444 2342

c) List all the student names with their membership numbers

SQL> select s.studname, m.memno from student s, membership m where m.studno=s.studno;

STUDNAME MEMNO------------- --------abhijeet 1001arun 1002arvind 1003ashish 1004ashwin 1005

d) List all the issues for the current date with student and Book names

SQL> select i.issno, s.studname, b.bookname from iss_rec I, membership m, student s, book b2 where i.memno=m.memno and m.studno=s.studno and i.issdate=to_char(sysdate);

ISSNO STUDNAME BOOKNAME------- ------------ ---------------13 arvind P&S

e) List the details of students who borrowed book whose author is CJDATE

SQL> select * from student where studno in(select studno from membership where memno in 2 (select memno from iss_rec where bookno in(select bookno from book where author=’CJDATE’)));

STUDNO STUDNAME---------- -------------

505 ashwin

f) Give a count of how many books have been bought by each student


Department of Computer Science and EngineeringSQL> select s.studno, count(i.bookno) from student s.membership m, book b, 2 iss_rec I where s.studno=m.studno and b.bookno=i.bookno group by s.studno;

STUDNO COUNT(I.BOOKNO)---------- -----------------------501 5502 5503 5504 5505 5

g) Give a list of books taken by student with stud_no as 5

SQL> select bookname from book where bookno in (select bookno from iss_rec where2 memno in(select memno from membership where3 studno in(select studno from student where studno=5)));

BOOKNAME-------------NT

h) List the book details which are issued as of today

SQL> delete from book where bookno in(select bookno from iss_rec where issdate=to_char(sysdate));delete from book where bookno in (select bookno from iss_rec where issdate=to_char(sysdate))

Errors Observed:

ERROR at line 1:ORA-02292: integrity constraint (SCOTT.SYS_C00840) violated – child record found

i) Create a view which lists out the iss_no, iss _date, stud_name, book name

j) Create a view which lists the daily issues-date wise for the last one week

Viva-Vice:

1. Describe the three levels of data abstraction?The are three levels of abstraction:


Department of Computer Science and Engineering Physical level: The lowest level of abstraction describes how data are stored. Logical level: The next higher level of abstraction, describes what data are

stored in database and what relationship among those data. View level: The highest level of abstraction describes only part of entire

database.2. Define the "integrity rules"

There are two Integrity rules. Entity Integrity: States that “Primary key cannot have NULL value” Referential Integrity: States that “Foreign Key can be either a NULL

value or should be Primary Key value of other relation.

3. What is extension and intension?Extension -

It is the number of tuples present in a table at any instance. This is time dependent.

Intension - It is a constant value that gives the name, structure of table and the

constraints laid on it.

4. What is System R? What are its two major subsystems?System R was designed and developed over a period of 1974-79 at IBM San

Jose Research Center. It is a prototype and its purpose was to demonstrate that it is possible to build a Relational System that can be used in a real life environment to solve real life problems, with performance at least comparable to that of existing system.

Its two subsystems are Research Storage System Relational Data System.

5. How is the data structure of System R different from the relational structure? Unlike Relational systems in System R

Domains are not supported Enforcement of candidate key uniqueness is optional Enforcement of entity integrity is optional Referential integrity is not enforced

6. What is Data Independence?Data independence means that “the application is independent of the

storage structure and access strategy of data”. In other words, The ability to modify the schema definition in one level should not affect the schema definition in the next higher level.

Two types of Data Independence: Physical Data Independence: Modification in physical level should not

affect the logical level. Logical Data Independence: Modification in logical level should affect

the view level. NOTE: Logical Data Independence is more difficult to achieve

7. What is a view? How it is related to data independence?


Department of Computer Science and EngineeringA view may be thought of as a virtual table, that is, a table that does not

really exist in its own right but is instead derived from one or more underlying base table. In other words, there is no stored file that direct represents the view instead a definition of view is stored in data dictionary.

Growth and restructuring of base tables is not reflected in views. Thus the view can insulate users from the effects of restructuring and growth in the database. Hence accounts for logical data independence.

3. Database Schema for a Employee-pay scenario

employee(emp_id : integer , emp_name: string)department(dept_id: integer, dept_name:string)paydetails(emp_id : integer, dept_id: integer, basic: integer, deductions: integer, additions: integer, DOJ: date)payroll(emp_id : integer, pay_date: date)

For the above schema, perform the following—a) Create the tables with the appropriate integrity constraintsb) Insert around 10 records in each of the tablesc) List the employee details department wised) List all the employee names who joined after particular date e) List the details of employees whose basic salary is between 10,000 and

20,000f) Give a count of how many employees are working in each departmentg) Give a names of the employees whose netsalary>10,000 h) List the details for an employee_id=5i) Create a view which lists out the emp_name, department, basic,

dedeuctions, netsalaryj) Create a view which lists the emp_name and his netsalary


Department of Computer Science and EngineeringAIM: Create the tables with the appropriate integrity constraints

Insert around 10 records in each of the tables


RAM : 256 MB

Hard Disk : 40 GB

Software : Oracle Create table employee(emp_id number(5) primary key,emp_name varchar2(25));

SQL>desc employee;

Name Null? Type………………………………………………………………………………………..EMP_ID NOT NULL NUMBER(5)EMP_NAME VARCHAR2(25)

Valid Test Data:

SQL>insert into employee values(&emp_id,’&emp_name’);SQL>select * from employee;

EMP_ID EMP_NAME………………………………………………………….

10 Robert21 Coulthard

30 Fernando Alonso 39 Kartikeyan 87 Kimmi

SQL>create table department(dept_id number(5) primary key,dept_name varchar2(20));

SQL>desc department;Name Null? Type………………………………………………………………………………………..DEPT_ID NOT NULL NUMBER(5)DEPT_NAME VARCHAR2(20)

SQL>insert into department values(&dept_id,’&dept_name’);SQL>select * from department;

DEPT_ID DEPT_NAME……………………………………………………………………………..

100 sales101 accounts


Department of Computer Science and Engineering102 administration103 production104 supervisor

SQL>create table paydetails(emp_id number(5) references employee(emp_id),dept_id number(5) reerences department(dept_id),basic number(7,2),deductions number(5,2),additions number(5,2),doj date);

SQL>desc paydetails;

Name Null? Type………………………………………………………………………………………..EMP_ID NUMBER(5)DEPT_ID NUMBER(5)BASIC NUMBER(7,2)DEDUCTIONS NUMBER(5,2)ADDITIONS NUMBER(5,2)DOJ DATE

Different Data Sets:SQL>insert into paydeatils values(&emp_id,&dept_id,&basic,&deductions,&additions,&doj);

SQL>select * from paydeatils;

EMP_ID DEPT_ID BASIC DEDUCTIONS ADDITIONS DOJ…………………………………………………………………………………………………………………..10 101 25023.12 43.09 71.23 08-JAN-9321 100 10500.29 23.98 40.9 01-JAN-06 30 102 6500.5 30.54 15 06-JUL-9739 103 9700.45 32.78 65.09 08-AUG-0387 104 15000 97.66 154.8 24-SEP-04

SQL>create table payroll(emp_id number(5)references employee(emp_id),pay_date date);

SQL>desc payroll;

Name Null? Type………………………………………………………………………………………..EMP_ID NUMBER(5)PAY_DATE DATE

SQL>insert into payroll values(&emp_id,’&date’);

SQL>select * from payroll;EMP_ID PAY_DATE………………………………………………………….10 31-JAN-06


Department of Computer Science and Engineering21 03-FEB-0630 15-JAN-0639 27-JAN-0687 04-FEB-06c) List the employee details department wise

SQL>select empid,deptid from paydet;

EMPID DEPTID…………………………401 500402 200403 600404 400405 1200

d) List all the employee names who joined after particular date

SQL>select e,empname from employee e,paydet p where e.empid=p.empid and p.doj>=’05-mar-06’;

EMPNAME…………………AVINASHNITINPHALGUN

e) List the details of employees whose basic salary is between 10,000 and 20,000

sqL> Select empid,empname from employee where salary between 10000 and 20000;

EMPID EMPNAME…………………………….

402 AKHILA403 aaaaaaaa

EMPID EMPNAME…………………………….

AKHILA

f) Give a count of how many employees are working in each departmentSQL>select count(empid),deptid from paydet group by deptid;

COUNT (EMPID) DEPTID………………………………………………………

1 200 1 400


Department of Computer Science and Engineering1 5001 6001 1200

g) Give a names of the employees whose netsalary>10,000

SQL> select empname from employee where empid in(select empid from paydet where basic-deduction>10000);

EMPNAME………………AVINASHAKHILAHARISHNITINPHALGUN

h) List the details for an employee_id=5

SQL> select * from employee where empid=5;

EMPID EMPNAME------------------------------------------5 Coulthard

i) Create a view which lists out the emp_name, department, basic, dedeuctions, netsalary

j) Create a view which lists the emp_name and his netsalary

Viva-Vice:13. What is Data Model?

A collection of conceptual tools for describing data, data relationships data semantics and constraints.

14. What is E-R model?This data model is based on real world that consists of basic objects called

entities and of relationship among these objects. Entities are described in a database by a set of attributes.

15. What is Object Oriented model?


Department of Computer Science and EngineeringThis model is based on collection of objects. An object contains values

stored in instance variables with in the object. An object also contains bodies of code that operate on the object. These bodies of code are called methods. Objects that contain same types of values and the same methods are grouped together into classes.

16. What is an Entity?It is a 'thing' in the real world with an independent existence.

17. What is an Entity type?It is a collection (set) of entities that have same attributes.

18. What is an Entity set?It is a collection of all entities of particular entity type in the database.


Department of Computer Science and Engineering4. Database Schema for a Video Library scenario

Customer(cust_no: integer,cust_name: string)Membership(Mem_no: integer, cust_no: integer)Cassette(cass_no:integer, cass_name:string, Language: String) Iss_rec(iss_no: integer, iss_date: date, mem_no: integer, cass_no: integer)

For the above schema, perform the following—a) Create the tables with the appropriate integrity constraintsb) Insert around 10 records in each of the tablesc) List all the customer names with their membership numbersd) List all the issues for the current date with the customer names and cassette

namese) List the details of the customer who has borrowed the cassette whose title is

“ The Legend”f) Give a count of how many cassettes have been borrowed by each customerg) Give a list of book which has been taken by the student with mem_no as 5h) List the cassettes issues for todayi) Create a view which lists outs the iss_no, iss_date, cust_name, cass_namej) Create a view which lists issues-date wise for the last one week


Department of Computer Science and EngineeringAIM: Create the tables with the appropriate integrity constraints Insert around 10 records in each of the tables


RAM : 256 MB

Hard Disk : 40 GB

Software : Oracle

SQL>create table customer(cust_no number(5) primary key,cust_name varchar2(20));

SQL>desc customer;

Name Null? Type……………………………………………………………………………………………………………..CUST_NO NOT NULL NUMBER(5)CUST_NAME VARCHAR2(20)

Valid Test Data:

SQL>insert into customer values(&cust_no,’&cust_name’);SQL>select * from customer;

CUST_NO CUST_NAME……………………………………………………………….

50 scott51 pandey52 varshney53 naidu54 bhimbra

SQL>create table membership(mem_no number(5) primary key,cust_no number(5) references customer(cust_no));

SQL>dsec membership;Name Null? Type………………………………………………………………………………………………………...MEM_NO NOT NULL NUMBER(5)CUST_NO NUMBER(5)

SQL>insert into memship values(&mem_no,&cust_no);SQL>select * from memship;

MEM_NO CUST_NOBalaji Institute of Technology and Science 39

Department of Computer Science and Engineering…………………………………………………920 50981 51897 52820 53928 54 SQL>create table cassette(cass_no number(5) primary key,Cass_name varchar2(15),language varchar2(15));

SQL>desc cassette;

Name Null? Type………………………………………………………………………………………..CASS_NO NOT NULL NUMBER(5)CASS_NAME VARCHAR2(15)LANGUAGE VARCHAR2(15)

SQL>insert into cassette values(&cass_no,’&cass_name’,’&language’);

SQL>select * from cassette;

CASS_NO CASS_NAME LANGUAGE………………………………………………………………………………………1 tagore telugu2 the lion king English3 anniyan tamil4 indra telugu5 lord of rings English

SQL>create table issu_rec(iss_no number(5) primary key,iss_date date,mem_no number(5)references memship(mem_no),cass_no number(5) references cassette(cass_no));

SQL>desc issu_rec;Name Null? Type………………………………………………………………………………………………………...ISS_NO NOT NULL NUMBER(5)ISS_DATE DATEMEM_NO NUMBER(5)CASS_NO NUMBER(5)

SQL>select * from issu_rec;

ISS_NO ISS_DATE MEM_NO CASS_NO……………………………………………………………………………………22 07-JAN-06 920 123 10-JAN-00 981 226 10-JAN-06 897 5


Department of Computer Science and Engineering3 01-JAN-06 820 434 31-DEC-05 928 3

c) List all the customer names with their membership numbersSQL>select c.custname,m.memno from customer1 c,membership1 m where

c.custno=m.custno;

CUSTNAME MEMNO……………….. ………………..NIKHIL 51VIVEK 52SHRAVAN 58VAMSI 57SHIVA 56

d) List all the issues for the current date with the customer names and cassette names

SQL>select i.issno,c.custname,cc.cassettename from customer1 c,membership1 m,cassette cc,issrec1 I where i.issdate=to_char(sysdate) and c.custno=m.custno and i.cassno=cc.cassno and i.memno=m.memno;

OutPut:

no rows selected.

e) List the details of the customer who has borrowed the cassette whose title is “ The Legend”

f) Give a count of how many cassettes have been borrowed by each customerg) Give a list of book which has been taken by the student with mem_no as 5h) List the cassettes issues for todayi) Create a view which lists outs the iss_no, iss_date, cust_name, cass_namej) Create a view which lists issues-date wise for the last one week

Viva-Vice:

19. What is an Extension of entity type?The collections of entities of a particular entity type are grouped together

into an entity set.

20. What is Weak Entity set?An entity set may not have sufficient attributes to form a primary key, and

its primary key compromises of its partial key and primary key of its parent entity, then it is said to be Weak Entity set.

21. What is an attribute?It is a particular property, which describes the entity.


Department of Computer Science and Engineering22. What is a Relation Schema and a Relation?

A relation Schema denoted by R(A1, A2, …, An) is made up of the relation name R and the list of attributes Ai that it contains. A relation is defined as a set of tuples. Let r be the relation which contains set tuples (t1, t2, t3, ..., tn). Each tuple is an ordered list of n-values t=(v1,v2, ..., vn).

23. What is degree of a Relation?It is the number of attribute of its relation schema.

24. What is Relationship?It is an association among two or more entities.

25. What is Relationship set? The collection (or set) of similar relationships.

5.Database Schema for a student-Lab scenario

Student(stud_no: integer, stud_name: string, class: string)Class(class: string, descrip: string)Lab(mach_no: integer, Lab_no: integer, description: String)Allotment(Stud_no: Integer, mach_no: integer, dayof week: string)

For the above schema, perform the following—a) Create the tables with the appropriate integrity constraintsb) Insert around 10 records in each of the tablesc) List all the machine allotments with the student names, lab and machine

numbersd) List the total number of lab allotments day wisee) Give a count of how many machines have been allocated to the ‘CSIT’ classf) Give a machine allotment etails of the stud_no 5 with his personal and class

detailsg) Count for how many machines have been allocatedin Lab_no 1 for the day

of the week as “Monday”h) How many students class wise have allocated machines in the labsi) Create a view which lists out the stud_no, stud_name, mach_no, lab_no,

dayofweekj) Create a view which lists the machine allotment details for “Thursday”


Department of Computer Science and EngineeringAIM: Create the tables with the appropriate integrity constraints

Insert around 10 records in each of the tables


RAM : 256 MB

Hard Disk : 40 GB

Software : Oracle

SQL>create table stu(stud_no number(5) primary key,stud_nam varchar2(20),class varchar2(20));

SQL> desc stu;

Name null? TypeSTUD_NO NOT NULL NUMBER(5)STUD_NAM VARCHAR2(20)CLASS VARCHAR2(20)

Valid Data Sets:

SQL> insert into stu values(&stud_no,’&stud_nam’,’&class’);SQL> select * from stu;

STUD_NO STUD_NAM CLASS39 LEON CSE34 VIKAS CSIT18 MATHEW ECE8 HANSEN MECH24 ALEXIS EEE

SQL> Create table class (class varchar2(20), descript varchar2(10));SQL> Describe class;

Name null type

CLASS VARCHAR2(10)DESCRIPT VARCHAR2(20)

SQL> create table lab(match_no number(5), lab_no number(5), description varchar2(20));

SQL> desc lab;

Name null type


Department of Computer Science and EngineeringMACH_NO NOT NULL NUMBER(5)LAB_NO NUMBER(5)DESCRIPTION VARCHAR2(20)

SQL> insert into lab values(&mach_no,&lab_no,’&description’);

SQL> select * from lab;

MATCH_NO LAB_NO DESCRIPTION--------------- --------- --------------------23 7 physics78 2 chemistry87 1 edc12 10 cds8 3 java lab

SQL> create table allotment(stud_no number(5) references stu(stud_no), match_no number(5) references lab(mach_no),Doweek varchar2(20));

SQL> desc allotment;

Name Null? Type-------------- ------- ---------STUD_NO NUMBER(5)MACH_NO NUMBER(5)DOWEEK VARCHAR2(20)

SQL>select * from allotment;

STUD_NO MACH_NO DOWEEK------------- -- ------------ ------------39 23 sat34 87 mon18 78 tue8 12 wed24 12 thu

c) List all the machine allotments with the student names, lab and machine numbers

SQL>select s.studname,l.machno from student1 s,lab l,allotment a where a.machno=l.machno and a.studno=s.studno;

STUDNAME MACHNO………………………………………..ABHIJEET 1


Department of Computer Science and EngineeringKALYAN 22 ASHWIN 3ARKA 4ARVIND 5

d) List the total number of lab allotments day wise

SQL>select l.machno,l.descrip,a.day from lab l,allotment a where a.machno=l.machno;

MACHNO DESCRIP DAY……………………………………………………………………1 UNIX MONDAY22 UNIX TUESDAY3 XP WEDNESDAY4 WINDOWS THRUSDAY5 ME FRIDAY

e) Give a count of how many machines have been allocated to the ‘CSIT’ class

SQL>select count(machno)from allotment where studno in(select studno from student1 where class=’CSIT’);

COUNT (MACHNO)…………………….. 1

f) Give a machine allotment etails of the stud_no 5 with his personal and class details

SQL>select a.studno,a.machno,s.studname,s.class from allotment a,student1 s where a.studno=s.studno and a.studno=503;

STUDNO MACHNO STUDNAME CLASS………………………………………………………………………………………………………503 5 ARVIND CSE

g) Count for how many machines have been allocatedin Lab_no 1 for the day of the week as “Monday”

h) How many students class wise have allocated machines in the labs

SQL>select count(studno) “allocated students in the labs”,class from student1 where studno in(select studno from allotment) group by class;

allocated students in the lab CLASS……………………………………………………………………………


Department of Computer Science and Engineering 2 CSE 1 ECE 1 EEE 1 IT

i) Create a view which lists out the stud_no, stud_name, mach_no, lab_no, dayofweek

j) Create a view which lists the machine allotment details for “Thursday”

Viva-Vice:

26. What is Relationship type? Relationship type defines a set of associations or a relationship set among

a given set of entity types. 27. What is degree of Relationship type?

It is the number of entity type participating.

25. What is DDL (Data Definition Language)?A data base schema is specifies by a set of definitions expressed by a

special language called DDL.

26. What is VDL (View Definition Language)?It specifies user views and their mappings to the conceptual schema.

27. What is SDL (Storage Definition Language)?This language is to specify the internal schema. This language may specify

the mapping between two schemas.

28. What is Data Storage - Definition Language?The storage structures and access methods used by database system are

specified by a set of definition in a special type of DDL called data storage-definition language.

29. What is DML (Data Manipulation Language)?This language that enable user to access or manipulate data as organised

by appropriate data model. Procedural DML or Low level: DML requires a user to specify what data are

needed and how to get those data. Non-Procedural DML or High level: DML requires a user to specify what data

are needed without specifying how to get those data.

31. What is DML Compiler?


Department of Computer Science and EngineeringIt translates DML statements in a query language into low-level instruction

that the query evaluation engine can understand.



6). Write a program to find largest number from the given three numbers.

Aim: To find largest number from the given three numbers.


RAM : 256 MB

Hard Disk : 40 GB

Software : Oracle, PlSQL

Algorithm:

Step 1: Declare the variable A, B, and C.Step 2: Store the valid data.Step 3: Compare variable A with B and A with CStep 4: If the value stored in variable A is big, it displays “A is Big”. (IF conditional statement should be used)Step 5: Compare variable B with CStep 6: If the value stored in variable B is big, it displays “B is Big”.Step 7: other wise it displays “C is Big”

Declare

A number;

B number;

C number;

Begin

A:=&a;

B:=&b;

C:=&c;

If a > b && a> c then

Dbms_output.put_line(‘ A is big ‘);

Else

If( b>c && b> a ) then


Department of Computer Science and EngineeringDbms_output.put_line(‘ B is big ‘);

Else

Dbms_output.put_line(‘ C is big ‘);

End if;

End if;

End;

Valid Data Sets:

Enter the value of a:1Enter the value of b:2Enter the value of c:3

OUTPUT:

C is big

Invalid Data sets :

Enter the value of a:yEnter the value of b:xEnter the value of c:a

Output:

Invalid data types.

Viva-Vice:

31. What is Pl-SQL ? Procedural Language Structured Query Language

32. What is Query evaluation engine?It executes low-level instruction generated by compiler.

33. What is DDL Interpreter?It interprets DDL statements and record them in tables containing

metadata.



34. What is Record-at-a-time?The Low level or Procedural DML can specify and retrieve each record from

a set of records. This retrieve of a record is said to be Record-at-a-time.

35. What is Set-at-a-time or Set-oriented?The High level or Non-procedural DML can specify and retrieve many

records in a single DML statement. This retrieve of a record is said to be Set-at-a-time or Set-oriented.

36. What is Relational Algebra?It is procedural query language. It consists of a set of operations that take

one or two relations as input and produce a new relation.

37. What is Relational Calculus?It is an applied predicate calculus specifically tailored for relational

databases proposed by E.F. Codd. E.g. of languages based on it are DSL ALPHA, QUEL.



7). Simple programs using loop, while and for iterative control statement.

a) To generate first 10 natural numbers using loop, while and for.

AIM: To generate first 10 natural numbers using loop, while and for.


RAM : 256 MB

Hard Disk : 40 GB


Algorithm:

Step 1: Declare the variable I.Step 2: Store the valid data 1 in I.Step 3: Use LOOP statementStep 4: Display the first value.Step 5: Increment the value of I by 1 value.Step 6: check the value up to 10 no. and repeat the loopStep 7: If condition exceeds the given value 10, the loop will be terminated.

/* using loop statement */

Declare

I number;

Begin

I:=1;

Loop

Dbms_output.put_line(I);

I:=I+1;

Exit when I>10;

End loop;

End;



Algorithm: for WHILE loop

Step 1: Declare the variable I.Step 2: Store the valid data 1 in I.Step 3: Use WHILE statementStep 4: Check the value of I with value 10.Step 5: if the value of I reached to 10 the loop will be terminatedStep 6: otherwise display value of IStep 7: increment the next value of I using I=I+1.

/* using while */

Declare

I number;

Begin

I:=1;

While (I<=10)

loop


I:=I+1;

End loop;

End;


Department of Computer Science and EngineeringAlgorithm:Step 1: Declare the variable I.Step 2: Store the value 1 in var. I.Step 3: Use For… LOOP statementStep 4: Display the first value of I.Step 5: Increment the value of I by 1 value.Step 6: check the value up to 10 no. and repeat the loopStep 7: if the loop exceeds the value 10 then the loop will be terminated.

/* using for loop*/

Begin

For I in 1..10

loop


End loop;

End;

Valid Test Data:

OUTPUT

12345678910



Viva-Vice:

38. How does Tuple-oriented relational calculus differ from domain-oriented relational calculus

The tuple-oriented calculus uses a tuple variables i.e., variable whose only permitted values are tuples of that relation. E.g. QUELThe domain-oriented calculus has domain variables i.e., variables that range over the underlying domains instead of over relation. E.g. ILL, DEDUCE.

39. What is normalization? It is a process of analysing the given relation schemas based on their

Functional Dependencies (FDs) and primary key to achieve the properties Minimizing redundancy Minimizing insertion, deletion and update anomalies.

40. What is Functional Dependency? A Functional dependency is denoted by X Y between two sets of

attributes X and Y that are subsets of R specifies a constraint on the possible tuple that can form a relation state r of R. The constraint is for any two tuples t1 and t2 in r if t1[X] = t2[X] then they have t1[Y] = t2[Y]. This means the value of X component of a tuple uniquely determines the value of component Y.

41. When is a functional dependency F said to be minimal? Every dependency in F has a single attribute for its right hand side. We cannot replace any dependency X A in F with a dependency Y A where Y

is a proper subset of X and still have a set of dependency that is equivalent to F.

We cannot remove any dependency from F and still have set of dependency that is equivalent to F.

42. What is Multivalued dependency?Multivalued dependency denoted by X Y specified on relation

schema R, where X and Y are both subsets of R, specifies the following constraint on any relation r of R: if two tuples t1 and t2 exist in r such that t1[X] = t2[X] then t3 and t4 should also exist in r with the following properties t3[x] = t4[X] = t1[X] = t2[X] t3[Y] = t1[Y] and t4[Y] = t2[Y] t3[Z] = t2[Z] and t4[Z] = t1[Z]

where [Z = (R-(X U Y)) ] 43. What is Lossless join property?

It guarantees that the spurious tuple generation does not occur with respect to relation schemas after decomposition.



8. Program to check whether given number is Armstrong or not.

AIM: to check whether given number is Armstrong or not.


RAM : 256 MB

Hard Disk : 40 GB

Software : Oracle, Pl-SQL

Algorithm:

Step 1: Declare the variable N, S, D and DUP.Step 2: Store the value in var. N and var. DUP..Step 3: check for the value of N, which is not equal to 0.Step 4: divide value stored in N by 10 and store it var. D. (D=n%10).Step 5: the reminder will be multiply 3 times and store it in Var. S. Step 6: The coefficient will be calculated using FLOOR function. And store it in var. N.Step 7: repeat the Steps 3, 4, 5, and 6 till loop will be terminated.Step 8: Check whether the stored value and calculated values are sameStep 9: if both the values are same, then display “The given number is

Armstrong”Step 10: Otherwise display “it is not Armstrong” and terminate the loop.

Declare

N number;S number;D number;

Begin

N:=&n;

S:=0;

While(n!=0)

Loop

D=n%10;S:=s+(D*D*D);N:=floor(n/10);


Department of Computer Science and EngineeringEnd loop;

If (DUP=S) then

DBMS_output.put_line(‘number is armstrong’);

Else

DBMS_output.put_line(‘number is not armstrong’);

End if;

End;

Test Valid Data Set:

Enter value of n

153

OUTPUT:

number is Armstrong.

Viva-Vice :

44. What is 1 NF (Normal Form)?The domain of attribute must include only atomic (simple, indivisible) values.

45. What is Fully Functional dependency? It is based on concept of full functional dependency. A functional dependency X Y is full

functional dependency if removal of any attribute A from X means that the dependency does not hold any more.

46. What is 2NF? A relation schema R is in 2NF if it is in 1NF and every non-prime attribute A in R is fully

functionally dependent on primary key.

47. What is 3NF?A relation schema R is in 3NF if it is in 2NF and for every FD X A either of the following is

true X is a Super-key of R. A is a prime attribute of R.

In other words, if every non prime attribute is non-transitively dependent on primary key.

48. What is BCNF (Boyce-Codd Normal Form)?A relation schema R is in BCNF if it is in 3NF and satisfies an additional constraint that for

every FD X A, X must be a candidate key.


Department of Computer Science and Engineering49. What is 4NF?

A relation schema R is said to be in 4NF if for every Multivalued dependency X Y that holds over R, one of following is true X is subset or equal to (or) XY = R. X is a super key.



9. Write a program to generate all prime numbers below 100.

AIM: to generate all prime numbers below 100.


RAM : 256 MB

Hard Disk : 40 GB


Declare

I number;

J number;

C number;

Begin

While(i<=100)

Loop

C:=0;J:=1;

While(j<=i)

Loop

If(floor(i%j)=0) then

C:= C+1;

End if;

J:=j+1;

End loop;

If(c=2) then

Dbms_output.put_line(i);


Department of Computer Science and EngineeringEnd if;

Endloop; End;

Valid Test Data

OUTPUT:

2

3

5

7

11

.

.

99

Viva-Vice:

50. What is 5NF?A Relation schema R is said to be 5NF if for every join dependency {R1, R2, ..., Rn} that

holds R, one the following is true Ri = R for some i. The join dependency is implied by the set of FD, over R in which the left side is key of R. 51. What is Domain-Key Normal Form?

A relation is said to be in DKNF if all constraints and dependencies that should hold on the the constraint can be enforced by simply enforcing the domain constraint and key constraint on the relation.

52. What are partial, alternate,, artificial, compound and natural key?

Partial Key:It is a set of attributes that can uniquely identify weak entities and that are related to same

owner entity. It is sometime called as Discriminator.

Alternate Key:All Candidate Keys excluding the Primary Key are known as Alternate Keys.

Artificial Key: If no obvious key, either stand alone or compound is available, then the last resort is to

simply create a key, by assigning a unique number to each record or occurrence. Then this is known as developing an artificial key.

Compound Key:


Department of Computer Science and EngineeringIf no single data element uniquely identifies occurrences within a construct, then

combining multiple elements to create a unique identifier for the construct is known as creating a compound key.

Natural Key:When one of the data elements stored within a construct is utilized as the primary key,

then it is called the natural key.

53. What is indexing and what are the different kinds of indexing?Indexing is a technique for determining how quickly specific data can be found.Types:

Binary search style indexing B-Tree indexing Inverted list indexing Memory resident table Table indexing



10. Write a program to demonstrate the GOTO statement.

AIM: to demonstrate the GOTO statement


RAM : 256 MB

Hard Disk : 40 GB


Declare

I number;

Begin

I:=1;

If(i>=0) then

GOTO here;

Else

Dbms_output.put_line( ‘ I is negative’);

End if;

<<here>>

Dbms_output.put_line( ‘ I is positive’);

End;

Valid Test Data

OUTPUT:

I is positive

Viva-Vice:

54. What is system catalog or catalog relation? How is better known as?


Department of Computer Science and EngineeringA RDBMS maintains a description of all the data that it contains, information about every

relation and index that it contains. This information is stored in a collection of relations maintained by the system called metadata. It is also called data dictionary.

55. What is meant by query optimization?The phase that identifies an efficient execution plan for evaluating a query that has the least

estimated cost is referred to as query optimization.

56. What is join dependency and inclusion dependency?Join Dependency:

A Join dependency is generalization of Multivalued dependency.A JD {R1, R2, ..., Rn} is said to hold over a relation R if R1, R2, R3, ..., Rn is a lossless-join decomposition of R . There is no set of sound and complete inference rules for JD.

Inclusion Dependency:An Inclusion Dependency is a statement of the form that some columns of a relation are

contained in other columns. A foreign key constraint is an example of inclusion dependency.

57. What is durability in DBMS?Once the DBMS informs the user that a transaction has successfully completed, its effects

should persist even if the system crashes before all its changes are reflected on disk. This property is called durability.

58. What do you mean by atomicity and aggregation?Atomicity:

Either all actions are carried out or none are. Users should not have to worry about the effect of incomplete transactions. DBMS ensures this by undoing the actions of incomplete transactions.

Aggregation:A concept which is used to model a relationship between a collection of entities and

relationships. It is used when we need to express a relationship among relationships.

59. What is a Phantom Deadlock?In distributed deadlock detection, the delay in propagating local information might cause the

deadlock detection algorithms to identify deadlocks that do not really exist. Such situations are called phantom deadlocks and they lead to unnecessary aborts.

60. What is a checkpoint and When does it occur?A Checkpoint is like a snapshot of the DBMS state. By taking checkpoints, the DBMS can

reduce the amount of work to be done during restart in the event of subsequent crashes.



11. Write a program to demonstrate %type and %rowtype attributes

AIM: to demonstrate %type and %rowtype attributes


RAM : 256 MB

Hard Disk : 40 GB


Declare

My_Empno emp.empno%type;My_Ename emp.ename%type;My_Emprow emp%rowtype;No number;

Begin

No:=&no;

Select empno,ename into my_empno,my_ename from emp where empno=no;

If(SQl%rowcount=1) then

Dbms_output.put_line(‘empno is’ || my_empno || ‘ename is ‘ || my_ename);

Else

Dbms_output.put_line( ‘error’);

End if;

Select * into my_emprow from emp where empno=no;

If(SQl%rowcount=1) then

Dbms_output.put_line(‘empno is’ || my_emprow.empno || ‘ename is ‘ || my_emprow.ename);

Else

Dbms_output.put_line( ‘error’);



End;

Valid Test Data

Enter the value for no:

7788

OUTPUT

empno is 7788 ename is vinay s.

empno is 7788 ename is vinay s.

Viva-Vice

61. What are the different phases of transaction?Different phases are

Analysis phase Redo Phase Undo phase

62. What do you mean by flat file database?It is a database in which there are no programs or user access languages. It has no cross-file

capabilities but is user-friendly and provides user-interface management.

63. What is "transparent DBMS"?It is one, which keeps its Physical Structure hidden from user.

64. Brief theory of Network, Hierarchical schemas and their propertiesNetwork schema uses a graph data structure to organize records example for such a database

management system is CTCG while a hierarchical schema uses a tree data structure example for such a system is IMS.

65. What is a query?A query with respect to DBMS relates to user commands that are used to interact with a data

base. The query language can be classified into data definition language and data manipulation language.

66. What do you mean by Correlated subquery?Subqueries, or nested queries, are used to bring back a set of rows to be used by the parent

query. Depending on how the subquery is written, it can be executed once for the parent query or it can be executed once for each row returned by the parent query. If the subquery is executed for each row of the parent, this is called a correlated subquery.

A correlated subquery can be easily identified if it contains any references to the parent subquery columns in its WHERE clause. Columns from the subquery cannot be referenced anywhere else in the parent query. The following example demonstrates a non-correlated subquery.


Department of Computer Science and Engineering E.g. Select * From CUST Where '10/03/1990' IN (Select ODATE From ORDER Where

CUST.CNUM = ORDER.CNUM)

67. What are the primitive operations common to all record management systems?Addition, deletion and modification.



12. Write a program to demonstrate predefined exceptions

AIM: to demonstrate predefined exceptions


RAM : 256 MB

Hard Disk : 40 GB


Declare

A number

B number;

C number;

Begin

A:=&a;

B:=&b;

C:=a/b;

Dbms_output.put_line(‘division is ‘ || C);

Exception

If (ZERO_DIVIDE) then

Dbms_output.put_line(‘b could not be zero’);

End if;

End;

Valid Test Data:

Enter the value for a:

10

Enter the value for b:


Department of Computer Science and Engineering0

OUTPUT:

b could not be zero

Viva-Vice:

68. Name the buffer in which all the commands that are typed in are stored‘Edit’ Buffer

69. What are the unary operations in Relational Algebra?PROJECTION and SELECTION.

70. Are the resulting relations of PRODUCT and JOIN operation the same?No.PRODUCT: Concatenation of every row in one relation with every row in another.JOIN: Concatenation of rows from one relation and related rows from another.

71. What is RDBMS KERNEL?Two important pieces of RDBMS architecture are the kernel, which is the software, and the

data dictionary, which consists of the system-level data structures used by the kernel to manage the database

You might think of an RDBMS as an operating system (or set of subsystems), designed specifically for controlling data access; its primary functions are storing, retrieving, and securing data. An RDBMS maintains its own list of authorized users and their associated privileges; manages memory caches and paging; controls locking for concurrent resource usage; dispatches and schedules user requests; and manages space usage within its table-space structures.72. Name the sub-systems of a RDBMS

I/O, Security, Language Processing, Process Control, Storage Management, Logging and Recovery, Distribution Control, Transaction Control, Memory Management, Lock Management

73. Which part of the RDBMS takes care of the data dictionary? HowData dictionary is a set of tables and database objects that is stored in a special area of the

database and maintained exclusively by the kernel.

74. What is the job of the information stored in data-dictionary?The information in the data dictionary validates the existence of the objects, provides access to

them, and maps the actual physical storage location.

14. Write a program to demonstrate user defined exceptions

AIM: to demonstrate user defined exceptions

HW/SW requirements:


Department of Computer Science and EngineeringProcessor : AMD Athelon ™ 1.67 GHz

RAM : 256 MB

Hard Disk : 40 GB


Declare

A number

B number;

C number;

Mydivide_zero EXCEPTION;

Begin

A:=&a;

B:=&b;

If(B=0) then

Raise Mydivide_zero;

else

C:=a/b;

Dbms_output.put_line(‘division is ‘ || C);

End if;

Exception

If (mydivide_zero) then

Dbms_output.put_line(‘b could not be zero’);

End if;

End;

Valid Test Data:

Enter the value for a:

10Balaji Institute of Technology and Science 68

Department of Computer Science and EngineeringEnter the value for b:

0

OUTPUT:

b could not be zero

Viva-Vice:

75. Not only RDBMS takes care of locating data it also determines an optimal access path to store or retrieve the data

76. How do you communicate with an RDBMS?You communicate with an RDBMS using Structured Query Language (SQL)

77. Define SQL and state the differences between SQL and other conventional programming Languages

SQL is a nonprocedural language that is designed specifically for data access operations on normalized relational database structures. The primary difference between SQL and other conventional programming languages is that SQL statements specify what data operations should be performed rather than how to perform them.

78. Name the three major set of files on disk that compose a database in OracleThere are three major sets of files on disk that compose a database. All the files are binary. These are

Database files Control files Redo logs

The most important of these are the database files where the actual data resides. The control files and the redo logs support the functioning of the architecture itself.

All three sets of files must be present, open, and available to Oracle for any data on the database to be useable. Without these files, you cannot access the database, and the database administrator might have to recover some or all of the database using a backup, if there is one.

79. What is an Oracle Instance?The Oracle system processes, also known as Oracle background processes, provide functions

for the user processes—functions that would otherwise be done by the user processes themselvesOracle database-wide system memory is known as the SGA, the system global area or shared

global area. The data and control structures in the SGA are shareable, and all the Oracle background processes and user processes can use them.

The combination of the SGA and the Oracle background processes is known as an Oracle instance

80. What are the four Oracle system processes that must always be up and running for the database to be useable

The four Oracle system processes that must always be up and running for the database to be useable include DBWR (Database Writer), LGWR (Log Writer), SMON (System Monitor), and PMON (Process Monitor).





15. Create a Cursor which update the salaries of an Employee as follows.

1. if sal<1000then update the salary to 1500.

2. if sal>=1000 and <2000 then update the salary to 2500.

3. if sal>=2000 and <=3000 then update the salary to 4000.

And also count the no.of records have been updated.*/


RAM : 256 MB

Hard Disk : 40 GB


Declare

Cursor my_cur is select empno,sal from emp;

Xno emp.empno%type;Xsal emp.sal%type;C number;

Begin

Open my_cur;

C:=0;

Loop

Fetch my_cur into xno,xsal;

If(xsal<1000) then

Update emp set sal=3000 where empno=xno;

C:=c+1;

Else if(xsal>=2000) and xsa<3000) then



Department of Computer Science and EngineeringC:=c+1;

End if;

End if;

Exit when my_cur%NOTFOUND ;

End loop;

Close my_cur;

Dbma_output.put_line(c||’records have been successfully updated’);

End;

Sql>@a.sql;

records have been successfully updated

pl/sql procedure successfully completed.

Valid Test Data

Before executing the cursor, the records in emp table as follows

Sql>select * from emp;

OUTPUT:

EMPNO ENAME JOB MGR HIREDATE SAL COMMD EPTNO

-----------------------------------------------------------------

7369 SMITH CLERK 7902 17-DEC-80 2000 20

7499 ALLEN SALESMAN 7698 20-FEB-81 1600 300 30

7521 WARD SALESMAN 7698 22-FEB-81 1250 500 30

EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO

-------- ---------- --------- ---------- --------- ---------- ----------------------------------------

7566 JONES MANAGER 7839 02-APR-81 2975 20

7654 MARTIN SALESMAN 7698 28-SEP-81 1250 1400 30

7698 BLAKE MANAGER 7839 01-MAY-81 2850 30


Department of Computer Science and Engineering…

….

…

14 rows selected.

Viva-Vice:

81. What are database files, control files and log files. How many of these files should a database have at least? Why?Database Files

The database files hold the actual data and are typically the largest in size. Depending on their sizes, the tables (and other objects) for all the user accounts can go in one database file—but that's not an ideal situation because it does not make the database structure very flexible for controlling access to storage for different users, putting the database on different disk drives, or backing up and restoring just part of the database.

You must have at least one database file but usually, more than one files are used. In terms of accessing and using the data in the tables and other objects, the number (or location) of the files is immaterial.

The database files are fixed in size and never grow bigger than the size at which they were created

Control Files The control files and redo logs support the rest of the architecture. Any database

must have at least one control file, although you typically have more than one to guard against loss.

The control file records the name of the database, the date and time it was created, the location of

the database and redo logs, and the synchronization information to ensure that all three sets of files

are always in step. Every time you add a new database or redo log file to the database, the

information is recorded in the control files.

Redo Logs Any database must have at least two redo logs. These are the journals for the database;

the redo logs record all changes to the user objects or system objects. If any type of failure occurs, the changes recorded in the redo logs can be used to bring the database to a consistent state without losing any committed transactions. In the case of non-data loss failure, Oracle can apply the information in the redo logs automatically without intervention from the DBA.

The redo log files are fixed in size and never grow dynamically from the size at which they were created.

82. What is ROWID?The ROWID is a unique database-wide physical address for every row on every table. Once

assigned (when the row is first inserted into the database), it never changes until the row is deleted or the table is dropped.


Department of Computer Science and EngineeringThe ROWID consists of the following three components, the combination of which uniquely

identifies the physical storage location of the row. Oracle database file number, which contains the block with the rows Oracle block address, which contains the row The row within the block (because each block can hold many rows) The ROWID is used internally in indexes as a quick means of retrieving rows with a particular

key value. Application developers also use it in SQL statements as a quick way to access a row once they know the ROWID

83. What is Oracle Block? Can two Oracle Blocks have the same address?Oracle "formats" the database files into a number of Oracle blocks when they are first created

—making it easier for the RDBMS software to manage the files and easier to read data into the memory areas.

The block size should be a multiple of the operating system block size. Regardless of the block size, the entire block is not available for holding data; Oracle takes up some space to manage the contents of the block. This block header has a minimum size, but it can grow.

These Oracle blocks are the smallest unit of storage. Increasing the Oracle block size can improve performance, but it should be done only when the database is first created.

Each Oracle block is numbered sequentially for each database file starting at 1. Two blocks can have the same block address if they are in different database files.

84. What is database Trigger?A database trigger is a PL/SQL block that can defined to automatically execute for insert,

update, and delete statements against a table. The trigger can e defined to execute once for the entire statement or once for every row that is inserted, updated, or deleted. For any one table, there are twelve events for which you can define database triggers. A database trigger can call database procedures that are also written in PL/SQL.

85. Name two utilities that Oracle provides, which are use for backup and recovery.Along with the RDBMS software, Oracle provides two utilities that you can use to back up and

restore the database. These utilities are Export and Import. The Export utility dumps the definitions and data for the specified part of the database to an

operating system binary file. The Import utility reads the file produced by an export, recreates the definitions of objects, and inserts the data

If Export and Import are used as a means of backing up and recovering the database, all the changes made to the database cannot be recovered since the export was performed. The best you can do is recover the database to the time when the export was last performed.



19. create a procedure which generate all the prime numbers below the given number and count the no.of prime numbers.


RAM : 256 MB

Hard Disk : 40 GB


Create or replace procedure prime_proc(n IN number,tot OUT number) as

i number;

c number;

j number;

Begin

i:=1;

tot:=0;

while(i<=n)

loop

j:=1;

c:=0;

while(j<=i)

loop

if(mod(I,j)=0) then

c:=c+1;

end if;

j:=j+1;


Department of Computer Science and Engineeringend loop;

if(c=2) then

dbms_output.put_line(i);

tot:=tot+1;

end if;

i:=i+1;

end loop;

end;

/

Sql>procedure created.

declare

t number;

begin

prime_proc(10,t);

dbms_output.put_line(‘the total prime no .are’||t);

end;

Valid Test Data:

sql>set serveroutput on

OUTPUT

sql>/

2

3

5

7

The total prime no.are 4


Department of Computer Science and EngineeringPl/sql procedure successfully completed.

Viva-Vice

86. Name two utilities that Oracle provides, which are use for backup and recovery.Along with the RDBMS software, Oracle provides two utilities that you can use to back up and

restore the database. These utilities are Export and Import. The Export utility dumps the definitions and data for the specified part of the database to an

operating system binary file. The Import utility reads the file produced by an export, recreates the definitions of objects, and inserts the data

If Export and Import are used as a means of backing up and recovering the database, all the changes made to the database cannot be recovered since the export was performed. The best you can do is recover the database to the time when the export was last performed.

87. What are stored-procedures? And what are the advantages of using them.Stored procedures are database objects that perform a user defined operation. A stored

procedure can have a set of compound SQL statements. A stored procedure executes the SQL commands and returns the result to the client. Stored procedures are used to reduce network traffic.

88. Tables derived from the ERD a) Are totally unnormalisedb) Are always in 1NFc) Can be further denormalisedd) May have multi-valued attributes

(b) Are always in 1NF

89. Spurious tuples may occur due to i. Bad normalization ii. Theta joins iii. Updating tables from join

a) i & ii b) ii & iiic) i & iii d) ii & iii

(a) i & iii because theta joins are joins made on keys that are not primary keys.90. A B C is a set of attributes. The functional dependency is as follows AB -> B AC -> C C -> B

a) is in 1NFb) is in 2NFc) is in 3NFd) is in BCNF

(a) is in 1NF since (AC)+ = { A, B, C} hence AC is the primary key. Since C B is a FD given, where neither C is a Key nor B is a prime attribute, this it is not in 3NF. Further B is not functionally dependent on key AC thus it is not in 2NF. Thus the given FDs is in 1NF.



18. create a procedure which updates the salaries of an employees as follows.

1.if sal<1000 then update the salry to 1500.

2.if sal>=1000 and <=2400 then update the salary to 2500.*/


RAM : 256 MB

Hard Disk : 40 GB


Create or replace procedure myproc as

Cursor my_cur is select empno,sal from emp;

Xno emp.empno%type;

Xsal emp.sal%type;

C number;

Begin

Open my_cur;

C:=0;

Loop

Fetch my_cur into xno,xsal;

If(xsal<1000) then


C:=c+1;

Else

Is(xsal>=1000 and xsal<=2400) then


C:=c+1;



End if;

Exit when my_cur%NOTFOUND;

End loop;

Close my_cur;

Dbms_output.put_line(c||’records have been successfully updated’);

End;

/

Valid Test Data:

Procedure created.

Sql>exec myproc;

OUTPUT:

Records have been successfully completed.

/* create function which add two given numbers. (Simple programs) */

Create or replace function add_fun(a number,b number) return

Number as

C number;

Begin

C:=a+b;

Return c;

End;

/

Function created.



/*add_fun specification*/

Declare

Result number;

Begin

Result:=add_fun(10,20);

Dbms_output.put_line(‘the sum of 10 and 20 is’||result);

End;

Sql>/

The sum of 10 and 20 is 30

Pl/sql procedure successfully completed.

/*create a function which count total no.of employees having salary less than 6000.*/

/*function body*/

Create or replace function count_emp(esal number)return number as

Cursor vin_cur as Select empno,sal from emp;

Xno emp.empno%type;

Xsal emp.sal%type;

C number;

Begin

Open vin_cur;

C:=0;

loop

fetch vin_cur into xno,xsal;


Department of Computer Science and Engineeringif(xsal<esal) then

c:=c+1;

end if;

exit when vin_cur%notfound;

end loop;

close vin_cur;

return c;

end;

/

Function created.

/*function specification*/

Declare

Ne number;

Xsal number;

Begin

Ne:=count_emp(xsal);

Dbms_output.put_line(xsal);

Dbma_output.put_line(‘there are ‘||ne||;employees’);

End;

/

OUTPUT

There are 8 employees.

Viva-Vice:

91. In mapping of ERD to DFD a) entities in ERD should correspond to an existing entity/store in DFDb) entity in DFD is converted to attributes of an entity in ERD


Department of Computer Science and Engineeringc) relations in ERD has 1 to 1 correspondence to processes in DFDd) relationships in ERD has 1 to 1 correspondence to flows in DFD

(a) entities in ERD should correspond to an existing entity/store in DFD

92. A dominant entity is the entitya) on the N side in a 1 : N relationshipb) on the 1 side in a 1 : N relationshipc) on either side in a 1 : 1 relationshipd) nothing to do with 1 : 1 or 1 : N relationship

(b) on the 1 side in a 1 : N relationship

93. Select 'NORTH', CUSTOMER From CUST_DTLS Where REGION = 'N' Order By CUSTOMER Union Select 'EAST', CUSTOMER From CUST_DTLS Where REGION = 'E' Order By CUSTOMERThe above is

a) Not an errorb) Error - the string in single quotes 'NORTH' and 'SOUTH'c) Error - the string should be in double quotesd) Error - ORDER BY clause

(d) Error - the ORDER BY clause. Since ORDER BY clause cannot be used in UNIONS

94. What is Storage Manager? It is a program module that provides the interface between the low-level data stored in

database, application programs and queries submitted to the system. 95. What is Buffer Manager?

It is a program module, which is responsible for fetching data from disk storage into main memory and deciding what data to be cache in memory.

96. What is Transaction Manager?It is a program module, which ensures that database, remains in a consistent state despite

system failures and concurrent transaction execution proceeds without conflicting.

97. What is File Manager?It is a program module, which manages the allocation of space on disk storage and data

structure used to represent information stored on a disk.

98. What is Authorization and Integrity manager?It is the program module, which tests for the satisfaction of integrity constraint and checks the

authority of user to access data. 99. What are stand-alone procedures?

Procedures that are not part of a package are known as stand-alone because they independently defined. A good example of a stand-alone procedure is one written in a SQL*Forms application. These types of procedures are not available for reference from other Oracle tools. Another limitation of stand-alone procedures is that they are compiled at run time, which slows execution.


Department of Computer Science and Engineering100. What are cursors give different types of cursors.

PL/SQL uses cursors for all database information accesses statements. The language supports the use two types of cursors Implicit Explicit

101. What is cold backup and hot backup (in case of Oracle)? Cold Backup:

It is copying the three sets of files (database files, redo logs, and control file) when the instance is shut down. This is a straight file copy, usually from the disk directly to tape. You must shut down the instance to guarantee a consistent copy.

If a cold backup is performed, the only option available in the event of data file loss is restoring all the files from the latest backup. All work performed on the database since the last backup is lost. Hot Backup:

Some sites (such as worldwide airline reservations systems) cannot shut down the database while making a backup copy of the files. The cold backup is not an available option.

So different means of backing up database must be used — the hot backup. Issue a SQL command to indicate to Oracle, on a tablespace-by-tablespace basis, that the files of the tablespace are to backed up. The users can continue to make full use of the files, including making changes to the data. Once the user has indicated that he/she wants to back up the tablespace files, he/she can use the operating system to copy those files to the desired backup destination.

The database must be running in ARCHIVELOG mode for the hot backup option. If a data loss failure does occur, the lost database files can be restored using the hot

backup and the online and offline redo logs created since the backup was done. The database is restored to the most consistent state without any loss of committed transactions.

102. What are Armstrong rules? How do we say that they are complete and/or soundThe well-known inference rules for FDs

Reflexive rule : If Y is subset or equal to X then X Y. Augmentation rule:

If X Y then XZ YZ. Transitive rule:

If {X Y, Y Z} then X Z. Decomposition rule : If X YZ then X Y. Union or Additive rule: If {X Y, X Z} then X YZ. Pseudo Transitive rule : If {X Y, WY Z} then WX Z.

Of these the first three are known as Amstrong Rules. They are sound because it is enough if a set of FDs satisfy these three. They are called complete because using these three rules we can generate the rest all inference rules.

103. How can you find the minimal key of relational schema?Minimal key is one which can identify each tuple of the given relation schema uniquely. For

finding the minimal key it is required to find the closure that is the set of all attributes that are dependent on any given set of attributes under the given set of functional dependency.

Algo. I Determining X+, closure for X, given set of FDs FBalaji Institute of Technology and Science 83

Department of Computer Science and Engineering1. Set X+ = X2. Set Old X+ = X+

3. For each FD Y Z in F and if Y belongs to X+ then add Z to X+

4. Repeat steps 2 and 3 until Old X+ = X+

Algo.II Determining minimal K for relation schema R, given set of FDs F1. Set K to R that is make K a set of all attributes in R2. For each attribute A in K

a. Compute (K – A)+ with respect to Fb. If (K – A)+ = R then set K = (K – A)+

104. What do you understand by dependency preservation?Given a relation R and a set of FDs F, dependency preservation states that the closure of

the union of the projection of F on each decomposed relation Ri is equal to the closure of F. i.e., ((R1(F)) U … U (Rn(F)))+ = F+

if decomposition is not dependency preserving, then some dependency is lost in the decomposition. 105. What is meant by Proactive, Retroactive and Simultaneous Update.

Proactive Update:The updates that are applied to database before it becomes effective in real

world .Retroactive Update:

The updates that are applied to database after it becomes effective in real world .Simulatneous Update:

The updates that are applied to database at the same time when it becomes effective in real world .

106. What are the different types of JOIN operations?Equi Join: This is the most common type of join which involves only equality

comparisions. The disadvantage in this type of join is that there



SQL Questions:

1. Which is the subset of SQL commands used to manipulate Oracle Database structures, including tables?

Data Definition Language (DDL)

2. What operator performs pattern matching?LIKE operator

3. What operator tests column for the absence of data?IS NULL operator

4. Which command executes the contents of a specified file? START <filename> or @<filename>

5. What is the parameter substitution symbol used with INSERT INTO command? &

6. Which command displays the SQL command in the SQL buffer, and then executes it? RUN

7. What are the wildcards used for pattern matching? _ for single character substitution and % for multi-character substitution

8. State true or false. EXISTS, SOME, ANY are operators in SQL. True

9. State true or false. !=, <>, ^= all denote the same operation. True

10. What are the privileges that can be granted on a table by a user to others?Insert, update, delete, select, references, index, execute, alter, all

11. What command is used to get back the privileges offered by the GRANT command? REVOKE

12. Which system tables contain information on privileges granted and privileges obtained? USER_TAB_PRIVS_MADE, USER_TAB_PRIVS_RECD

13. Which system table contains information on constraints on all the tables created? USER_CONSTRAINTS

14. TRUNCATE TABLE EMP;DELETE FROM EMP;

Will the outputs of the above two commands differ? Both will result in deleting all the rows in the table EMP.

15. What is the difference between TRUNCATE and DELETE commands?


Department of Computer Science and Engineering TRUNCATE is a DDL command whereas DELETE is a DML command. Hence DELETE

operation can be rolled back, but TRUNCATE operation cannot be rolled back. WHERE clause can be used with DELETE and not with TRUNCATE.

16. What command is used to create a table by copying the structure of another table?Answer :

CREATE TABLE .. AS SELECT commandExplanation :

To copy only the structure, the WHERE clause of the SELECT command should contain a FALSE statement as in the following.

CREATE TABLE NEWTABLE AS SELECT * FROM EXISTINGTABLE WHERE 1=2;If the WHERE condition is true, then all the rows or rows satisfying the condition will be

copied to the new table.

17. What will be the output of the following query?SELECT REPLACE(TRANSLATE(LTRIM(RTRIM('!! ATHEN !!','!'), '!'), 'AN', '**'),'*','TROUBLE') FROM DUAL;

TROUBLETHETROUBLE

18. What will be the output of the following query?SELECT DECODE(TRANSLATE('A','1234567890','1111111111'), '1','YES', 'NO' );Answer :

NOExplanation :

The query checks whether a given string is a numerical digit.

19. What does the following query do?SELECT SAL + NVL(COMM,0) FROM EMP; This displays the total salary of all employees. The null values in the commission column will

be replaced by 0 and added to salary.

20. Which date function is used to find the difference between two dates? MONTHS_BETWEEN

21. Why does the following command give a compilation error?DROP TABLE &TABLE_NAME; Variable names should start with an alphabet. Here the table name starts with an '&' symbol.

22. What is the advantage of specifying WITH GRANT OPTION in the GRANT command? The privilege receiver can further grant the privileges he/she has obtained from the owner to

any other user.

23. What is the use of the DROP option in the ALTER TABLE command? It is used to drop constraints specified on the table.

24. What is the value of ‘comm’ and ‘sal’ after executing the following query if the initial value of ‘sal’ is 10000?

UPDATE EMP SET SAL = SAL + 1000, COMM = SAL*0.1; sal = 11000, comm = 1000



25. What is the use of DESC in SQL?Answer :

DESC has two purposes. It is used to describe a schema as well as to retrieve rows from table in descending order.Explanation :

The query SELECT * FROM EMP ORDER BY ENAME DESC will display the output sorted on ENAME in descending order.

26. What is the use of CASCADE CONSTRAINTS? When this clause is used with the DROP command, a parent table can be dropped even when a

child table exists.

27. Which function is used to find the largest integer less than or equal to a specific value? FLOOR

28. What is the output of the following query?SELECT TRUNC(1234.5678,-2) FROM DUAL; 1200



REFERENCES:

SCOTT Urman, oracle 8i-PL/SQL programming,TMH-2000 Loney, oracle 8i-the complete reference,TMH-2000 Loney, oracle 9i-the complete reference,TMH-2000 Bayross, oracle Teach Your Self SQL/PLSQL using oracle 8i and 9i with

SQLJ, BPB, 2002. Abbey, oracle 8i-A beginners guide,TMH-2000.



GURU NANAK Engineering CollegeIbrahimpatnam, R R District – 501 506 (A. P.)

Assignment / Record No - _____

Department of ________________________

Name : ______________________________________

Class / Year / Semester

: B. Tech _____ Year _________ Semester

Roll No. : _________________________

Subject : _________________________

Branch : _________________________

Date of Submission

: _____ / _____ / __________

________________ ________________Student Signature Faculty Signature

Index Page

S. No. Date

Program / Experiment Page No.

Sign / Remarks

Balaji Institute of Technology and Science

Annexure - 1

Annexure - 2

89


GURU NANAK Engineering CollegeIbrahimpatnam, R R District – 501 506 (A. P.)

(Sponsored by: Guru Nanak Educational Society, Hyderabad (A. P.)

Balaji Institute of Technology and Science

Annexure - 3

90


Department of ________________________

This is to certify that

Name : ______________________________________

Class / Year / Semester

: B. Tech ______ Year ________ Semester

Roll No. : _________________________

Subject : _________________________

Branch : _________________________

Has successfully completed the course of programs / experiments prescribed by Jawharlal Nerhu

Technological Univeristy of the department during the academic year 200__ - 200 __.

________________ ________________Faculty Incharge Department Head


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