DBMS_Lab_10CSL57_2012

10CSL57 SRSIT/CSE/DBA/V.0.1

DATABASE APPLICATIONS LABORATORY(COMMON TO CSE)

Subject Code: 10CSL57

1. Consider the following relations:

Student (snum: integer, sname: string, major: string, level: string, age: integer)

Class (name: string, meets at: string, room: string, d: integer)

Enrolled (snum: integer, cname: string)

Faculty (fid: integer, fname: string, deptid: integer)

The meaning of these relations is straightforward; for example, Enrolled has one record

per student-class pair such that the student is enrolled in the class. Level is a two

character code with 4 different values (example: Junior: JR etc)

Write the following queries in SQL. No duplicates should be printed in any of the

answers.

i. Find the names of all Juniors (level = JR) who are enrolled in a class

taught by Prof. Harshith

ii. Find the names of all classes that either meet in room R128 or have five or

more Students enrolled.

iii. Find the names of all students who are enrolled in two classes that meet at

the same time.

iv. Find the names of faculty members who teach in every room in which

some class is taught.

v. Find the names of faculty members for whom the combined enrollment of

the courses that they teach is less than five.

2. The following relations keep track of airline flight information:

Flights (no: integer, from: string, to: string, distance: integer, Departs: time,

arrives: time, price: real)

Aircraft (aid: integer, aname: string, cruisingrange: integer)

Certified (eid: integer, aid: integer)

Employees (eid: integer, ename: string, salary: integer)

CM/SB/S Page 1 2012-2013


Note that the Employees relation describes pilots and other kinds of employees as well;

every pilot is certified for some aircraft, and only pilots are certified to fly.

Write each of the following queries in SQL.

i. Find the names of aircraft such that all pilots certified to operate them have

salaries more than Rs.80, 000.

ii. For each pilot who is certified for more than three aircrafts, find the eid and

the maximum cruisingrange of the aircraft for which she or he is certified.

iii. Find the names of pilots whose salary is less than the price of the cheapest

route from Bengaluru to Frankfurt.

iv. For all aircraft with cruisingrange over 1000 Kms, .find the name of the

aircraft and the average salary of all pilots certified for this aircraft.

v. Find the names of pilots certified for some Boeing aircraft.

vi. Find the aids of all aircraft that can be used on routes from Bengaluru to

New Delhi.

3. Consider the following database of student enrollment in courses & books

adopted for each course.

STUDENT (regno: string, name: string, major: string, bdate:date)

COURSE (course #:int, cname:string, dept:string)

ENROLL ( regno:string, course#:int, sem:int, marks:int)

BOOK _ ADOPTION (course# :int, sem:int, book-ISBN:int)

TEXT (book-ISBN:int, book-title:string, publisher:string, author:string)

i. Create the above tables by properly specifying the primary keys and the

foreign keys.

ii. Enter at least five tuples for each relation.

iii. Demonstrate how you add a new text book to the database and make this

book be adopted by some department.

iv. Produce a list of text books (include Course #, Book-ISBN, Book-title) in

the alphabetical order for courses offered by the ‘CS’ department that use

more than two books.

CM/SB/S Page 2 2012-2013


v. List any department that has all its adopted books published by a specific

publisher.

vi. Generate suitable reports.

vii. Create suitable front end for querying and displaying the results.

4. The following tables are maintained by a book dealer.

AUTHOR (author-id:int, name:string, city:string, country:string)

PUBLISHER (publisher-id:int, name:string, city:string, country:string)

CATALOG (book-id:int, title:string, author-id:int, publisher-id:int, category-

id:int, year:int, price:int)

CATEGORY (category-id:int, description:string)

ORDER-DETAILS (order-no:int, book-id:int, quantity:int)

i. Create the above tables by properly specifying the primary keys and

the foreign keys.


iii. Give the details of the authors who have 2 or more books in the

catalog and the price of the books is greater than the average price of

the books in the catalog and the year of publication is after 2000.

iv. Find the author of the book which has maximum sales.

v. Demonstrate how you increase the price of books published by a

specific publisher by 10%.



5. Consider the following database for a banking enterprise

BRANCH(branch-name:string, branch-city:string, assets:real)

ACCOUNT(accno:int, branch-name:string, balance:real)

DEPOSITOR(customer-name:string, accno:int)

CUSTOMER(customer-name:string, customer-street:string, customer-city:string)

LOAN(loan-number:int, branch-name:string, amount:real)

BORROWER(customer-name:string, loan-number:int)


the foreign keys

CM/SB/S Page 3 2012-2013


ii. Enter at least five tuples for each relation

iii. Find all the customers who have at least two accounts at the Main

branch.

iv. Find all the customers who have an account at all the branches located

in a specific city.

v. Demonstrate how you delete all account tuples at every branch located

in a specific city.



Instructions:

1. The exercises are to be solved in an RDBMS environment like Oracle or DB2.

2. Suitable tuples have to be entered so that queries are executed correctly.

3. Front end may be created using either VB or VAJ or any other similar tool.

4. The student need not create the front end in the examination. The results of the

queries may be displayed directly.

5. Relevant queries other than the ones listed along with the exercises may also be

asked in the examination.

6. Questions must be asked based on lots.

CM/SB/S Page 4 2012-2013


PROBLEM 1

1. Consider the following relations:

Student (snum: integer, sname: string, major: string, level: string, age: integer)

Class (name: string, meets at: string, room: string, d: integer)

Enrolled (snum: integer, cname: string)

Faculty (fid: integer, fname: string, deptid: integer)

The meaning of these relations is straightforward; for example, Enrolled has one record

per student-class pair such that the student is enrolled in the class. Level is a two

character code with 4 different values (example: Junior: JR etc)

Write the following queries in SQL. No duplicates should be printed in any of the

answers.

vi. Find the names of all Juniors (level = JR) who are enrolled in a class

taught by Prof. Harshith

vii. Find the names of all classes that either meet in room R128 or have five or

more Students enrolled.

viii. Find the names of all students who are enrolled in two classes that meet at

the same time.

ix. Find the names of faculty members who teach in every room in which

some class is taught.

Find the names of faculty members for whom the combined enrollment of the courses

that they teach is less than five

TABLE CREATION

SQL> create table student( snum int not null primary key, sname varchar(20), major varchar(20), level varchar(10), age int not null );

Table created.

CM/SB/S Page 5 2012-2013


INSERTING TUPLES

SQL> insert into student values(01,'raju','ise','JR',20);Enter value for snum: 01Enter value fro sname:rajuEnter value for major: iseEnter value for level: JREnter value for age: 20 old 1: insert into student values(‘&snum’,’&sname’,’&major’,’&level’,’&age’);new 1: insert into student values(01,'raju','ise','JR',20);

1 row created

SQL> insert into student values(02,'ram','ise','JR',20);SQL> insert into student values(03,'akash','ise','SR',21);SQL> insert into student values(04,'akash','cse','SR',21);SQL> insert into student values(05,'avinash','ise','FR',19);SQL> insert into student values(06,'kumar','ise','FR',19);SQL> insert into student values(07,'john','ise','SSR',22);

SQL> select * from student;+------+---------+-------+-------+-----+| snum | sname | major | level | age |+------+---------+-------+-------+-----+| 1 | raju | ise | JR | 20 | | 2 | ram | ise | JR | 20 | | 3 | akash | ise | SR | 21 | | 4 | akash | cse | SR | 21 | | 5 | avinash | ise | FR | 19 | | 6 | kumar | ise | FR | 19 | | 7 | john | ise | SSR | 22 | +------+---------+-------+-------+-----+

TABLE CREATION

SQL> create table class( name varchar(10) primary key, meets_at varchar(10), room varchar(10), fid int not null);

Table created

INSERTING TUPLES

CM/SB/S Page 6 2012-2013


SQL> insert into class values('3','hall1','101',1);Enter value for name: 3Enter value fro meets_at : hall1Enter value for room: 101Enter value for fid: 1 old 1: insert into class values(‘&name’,’&meets_at’,’&room’,’&fid’);new 1: insert into class values('3','hall1','101',1);

1 row created

SQL> insert into class values('4','hall1','101',1);SQL> insert into class values('5','hall2','102',2);SQL> insert into class values('6','hall2','102',1);SQL> insert into class values('7','hall2','103',3);SQL> insert into class values('8','hall3','103',3);SQL> insert into class values('1','hall3','103',3);

SQL> select * from class;+------+----------+------+-----+| name | meets_at | room | fid |+------+----------+------+-----+| 3 | hall1 | 101 | 1 | | 4 | hall1 | 101 | 1 | | 5 | hall2 | 102 | 2 | | 6 | hall2 | 102 | 1 | | 7 | hall2 | 103 | 3 | | 8 | hall3 | 103 | 3 | | 1 | hall3 | 103 | 3 | +------+----------+------+-----+

TABLE CREATIONSQL> create table enrolled( snum int not null, cname varchar(10), primary key(snum,cname), foreign key(snum) references student(snum), foreign key(cname) references class(name));Table Created

INSERTING TUPLES

SQL> insert into enrolled values(1,3);Enter value for snum: 1Enter value for cname : 3old 1: insert into class values(‘&snum’,’&cname’);

CM/SB/S Page 7 2012-2013


new 1: insert into class values(1,3);

1 row created

SQL> insert into enrolled values(1,4);SQL> insert into enrolled values(1,5);SQL> insert into enrolled values(2,5);SQL> insert into enrolled values(3,5);SQL> insert into enrolled values(4,5);SQL> insert into enrolled values(4,1);

SQL> select * from enrolled;+------+-------+-----+| snum | cname |+------+-------+| 1 | 3 | | 1 | 4 | | 1 | 5 | | 2 | 5 | | 3 | 5 | | 4 | 1 | | 4 | 5 | +------+-------+

TABLE CREATION

SQL> create table faculty( fid int not null primary key, fname varchar(25), deptid int not null, foreign key(fid) references class(fid));Table created

INSERTING TUPLES

SQL> insert into faculty values(1,'Prof Harshit',200);Enter value for fid: 1Enter value for fname : Prof HarshitEnter values for deptid: 200old 1: insert into faculty values(‘&fid’,’&fname’,’&deptid’);new 1: insert into class values(1,3);1 row created SQL> insert into faculty values(2,'Prof Harshit',200);SQL> insert into faculty values(3,'Prof Raghav',201);SQL> insert into faculty values(4,'Prof Raghav',201);

CM/SB/S Page 8 2012-2013


SQL> insert into faculty values(5,'Prof Pranav',202);SQL> insert into faculty values(6,'Prof Pranav',202);

SQL> select * from faculty;+-----+--------------+--------+| fid | fname | deptid |+-----+--------------+--------+| 1 | Prof Harshit | 200 | | 2 | Prof Harshit | 200 | | 3 | Prof Raghav | 201 | | 4 | Prof Raghav | 201 | | 5 | Prof Pranav | 202 | | 6 | Prof Pranav | 202 | +-----+--------------+--------+

QUERIES

i. Find the names of all Juniors (level = JR) who are enrolled in a class taught by Prof. Harshith

SQL> select distinct s.sname from student s,class c, enrolled e, faculty f where s.snum=e.snum and e.cname=c.name and c.fid=f.fid and f.fname='Prof Harshit' and s.level='JR';

RESULT+-------+| sname |+-------+| raju || ram |+-------+

ii. Find the names of all classes that either meet in room 101 or have five or more Students enrolled.

SQL> select c.name from class c where c.room='101' or c.name in(select e.cname from enrolled e group by e.cname having count(*)>=5);RESULT name --------------------- 3 4

iii. Find the names of all students who are enrolled in two classes that meet at the same time.

SQL> select distinct s.sname from student s where s.snum in(select a.snum from enrolled a, enrolled b, class a1, class b1

CM/SB/S Page 9 2012-2013


where a.snum=a.snum and a.cname<>b.cname and a.cname=a1.name and b.cname=b1.name and a1.meets_at=b1.meets_at);RESULT+-------+| sname |+-------+| raju | +-------+

iv. Find the names of faculty members who teach in every room in which some class is taught

SQL> select distinct f.fname from faculty f where not exists(select c.room from class c where c.fid=f.fid);

RESULT+-------------+| fname |+-------------+| Prof Raghav | | Prof Pranav | +-------------+

v. Find the names of faculty members for whom the combined enrollment of the courses that they teach is less than five.

SQL> select distinct f.fname from faculty f where 5 >(select count(e.snum) from class c, enrolled e where c.name=e.cname and c.fid=f.fid);RESULT+--------------+| fname |+--------------+| Prof Harshit | | Prof Raghav | | Prof Pranav | +--------------+

CM/SB/S Page 10 2012-2013


PROBLEM-2

2. The following relations keep track of airline flight information:

Flights (no: integer, from: string, to: string, distance: integer, Departs: time,

arrives: time, price: real)

Aircraft (aid: integer, aname: string, cruisingrange: integer)

Certified (eid: integer, aid: integer)

Employees (eid: integer, ename: string, salary: integer)

Note that the Employees relation describes pilots and other kinds of employees as well;

every pilot is certified for some aircraft, and only pilots are certified to fly.

Write each of the following queries in SQL.

i. Find the names of aircraft such that all pilots certified to operate them have

salaries more than Rs.80, 000.

ii. For each pilot who is certified for more than three aircrafts, find the eid and

the maximum cruisingrange of the aircraft for which she or he is certified.

iii. Find the names of pilots whose salary is less than the price of the cheapest

route from Bengaluru to Frankfurt.

iv. For all aircraft with cruisingrange over 1000 Kms, .find the name of the

aircraft and the average salary of all pilots certified for this aircraft.

v. Find the names of pilots certified for some Boeing aircraft.

vi. Find the aids of all aircraft that can be used on routes from Bengaluru to

New Delhi.

TABLE CREATION

SQL> create table flights( flight_no int not null primary key, flight_from varchar(20), flight_to varchar(20), distance int, departure datetime null, arrival datetime null, price decimal(10,2));Table createdINSERTING TUPLES

CM/SB/S Page 11 2012-2013


SQL> insert into flights values(7,'bangalore','mumbai',500,'2012-12-06 22:34:46', 3, 5000);1 row created

SQL> select * from flights;

TABLE CREATION

SQL> create table aircraft( aid int not null primary key, aname varchar(20), cruisingrange int null);Table Created

INSERTING TUPLES

SQL> insert into aircraft values(117,'kingfisher',10000);

1 row created

SQL> select * from aircraft;

TABLE CREATION

SQL> create table certified(

CM/SB/S Page 12 2012-2013


eid int not null, aid int not null, primary key(eid,aid), foreign key(aid) references aircraft(aid));Table Created

INSERTING TUPLESSQL> insert into certified values(109,117);1 row created

SQL> select * from certified;

TABLE CREATION

SQL> create table employees( eid int not null, ename varchar(20), salary int not null, primary key(eid), foreign key(eid) references certified(eid));Table Created

INSERTING TUPLES

SQL>insert into employees values(100,’pranav’,20000);1 row createdSQL> select * from employees;

QUERIES

i. Find the names of aircraft such that all pilots certified to operate them have salaries more than Rs.80, 000.

CM/SB/S Page 13 2012-2013


SQL> select distinct a.aname from aircraft a where a.aid in (select c.aid from certified c, employees e where c.eid=e.eid and not exists(select * from employees e1 where e1.eid=e.eid and e.salary<80000));RESULT+------------+| aname |+------------+| indian | | kingfisher | +------------+

ii. For each pilot who is certified for more than three aircrafts, find the eid and the maximum cruisingrange of the aircraft for which she or he is certified.

SQL> select c.eid,max(a.cruisingrange) from certified c, aircraft a where c.aid=a.aid group by c.eid having count(*)>3;

RESULT+-----+----------------------+| eid | max(a.cruisingrange) |+-----+----------------------+| 115 | 10000 | +-----+----------------------+

iii. Find the names of pilots whose salary is less than the price of the cheapest route from Bengaluru to Frankfurt.

SQL> select distinct e.ename from employees e where e.salary < (select min(f.price) from flights f where f.flight_from='bangalore' and f.flight_to='Frankfurt');

RESULT+----------+| ename |+----------+| pranav | | pranav p |

CM/SB/S Page 14 2012-2013


| ramesh | | rajiv | | shashank | | suraj | | boeing | +----------+

iv. For all aircraft with cruisingrange over 1000 Kms, .find the name of the aircraft and the average salary of all pilots certified for this aircraft.

SQL> select temp.name, temp.avgsalary from(select a.aid, a.aname as name,avg(e.salary) as avgsalary from aircraft a, certified c, employees e where a.aid=c.aid and c.eid=e.eid and a.cruisingrange>1000 group by a.aid, a.aname) as temp;

RESULT+--------+------------+| name | avgsalary |+--------+------------+| sahara | 30000.0000 | | sahara | 30000.0000 | | indian | 30000.0000 | | indian | 65000.0000 |

v. Find the names of pilots certified for some Kingfisher aircraft.

SQL> select distinct e.ename from employees e, certified c, aircraft a where e.eid=c.eid and c.aid=a.aid and a.aname like'kingfisher%';RESULT+----------+| ename |+----------+| pranav | | pranav p | | ramesh | | boeing | +----------+

vi. Find the aids of all aircraft that can be used on routes from Bengaluru to New Delhi.

SQL> select a.aid from aircraft a where a.cruisingrange>(select min(f.distance) from flights f

CM/SB/S Page 15 2012-2013


where f.flight_from='bangalore' and f.flight_to='mumbai');RESULT+-----+| aid |+-----+| 111 | | 112 | | 113 | | 114 | | 115 | | 116 | | 117 | | 118 | | 119 | | 120 | | 121 | | 122 | | 123 | +-----+

CM/SB/S Page 16 2012-2013


PROBLEM-3

3. Consider the following database of student enrollment in courses & books

adopted for each course.

STUDENT (regno: string, name: string, major: string, bdate:date)

COURSE (course #:int, cname:string, dept:string)

ENROLL ( regno:string, course#:int, sem:int, marks:int)

BOOK _ ADOPTION (course# :int, sem:int, book-ISBN:int)

TEXT (book-ISBN:int, book-title:string, publisher:string, author:string)

i. Create the above tables by properly specifying the primary keys and the

foreign keys.


iii. Demonstrate how you add a new text book to the database and make this book

be adopted by some department.

iv. Produce a list of text books (include Course #, Book-ISBN, Book-title) in the

alphabetical order for courses offered by the ‘CS’ department that use more

than two books.

v. List any department that has all its adopted books published by a specific

publisher.



TABLE CREATION

SQL> create table student ( regno varchar(10) primary key, name varchar(15) , major varchar(30) , bdate date );

Table created.

INSERTING TUPLES

SQL> insert into student values ('&regno','&name','&major','&bdate');SQL> insert into student values ('&regno','&name','&major','&bdate');

CM/SB/S Page 17 2012-2013


Enter value for regno: 101Enter value for name: akashEnter value for major: dbmsEnter value for bdate: 11-jan-1989old 2: ('&regno','&name','&major','&bdate')new 2: ('101','akash','dbms','11-jan-1989')

1 row created.

SQL> select * from student;

REGNO NAME MAJOR BDATE----------------------------------------------------------------------------101 akash dbms 11-JAN-89102 arvind algorithms 22-FEB-88103 anveesh unix 23-MAR-89104 virinchi game theory 24-APR-90105 vageesh game theory 25-MAY-90

TABLE CREATION

SQL> create table course ( course integer primary key, cname varchar(30) , dept varchar(30) );

Table created.

INSERTING TUPLES

SQL> insert into course values ('&course','&cname','&dept');Enter value for course: 1Enter value for cname: mtechEnter value for dept: cseold 2: ('&course','&cname','&dept')new 2: ('001','mtech','cse')

1 row created

SQL> select * from course;

COURSE CNAME DEPT---------------------------------------------------------------------- 1 mtech cse 2 me cse 3 mtech ise 4 bsc cse

CM/SB/S Page 18 2012-2013


5 mca ise

TABLE CREATION

SQL> create table text ( book_isbn integer primary key, book_title varchar(30) not null, publisher varchar(30) not null, author varchar(30) not null );

Table created.

INSERTING TUPLES

SQL>insert into text values ('&book_isbn','&book_title','&publisher','&author');Enter value for book_isbn: 111Enter value for book_title: linuxEnter value for publisher: pearsonEnter value for author: sumithaba dasold 2: ('&book_isbn','&book_title','&publisher','&author')new 2: ('111','linux','pearson','sumithaba das')

1 row created.

SQL> select * from text;

BOOK_ISBN BOOK_TITLE PUBLISHER----------------------------------------------------------------------------------------AUTHOR------------------------------ 111 linux pearsonsumithaba das

222 compiler design tmhtenenbaum 333 computer graphics pearsonmicheal folk

444 data structures venugopaltmh

555 file structures godsehimalaya

TABLE CREATION

SQL> create table enroll

CM/SB/S Page 19 2012-2013


( regno varchar(10) not null, course integer not null, sem integer not null, marks integer not null, primary key(regno,course,sem), foreign key (regno) references student (regno) on delete cascade, foreign key (course) references course (course) on delete cascade );

Table created.

INSERTING TUPLES

SQL>insert into enroll values ('&regno','&course','&sem','&marks');Enter value for regno: 103Enter value for course: 3Enter value for sem: 5Enter value for marks: 590old 2: ('&regno','&course','&sem','&marks')new 2: ('103','003','5','590')

SQL> select * from enroll;

REGNO COURSE SEM MARKS----------------------------------------------------103 3 5 590105 5 6 789102 1 4 678101 1 6 430104 4 4 556

TABLE CREATION

SQL> create table book_adoption ( course integer not null, sem integer not null, book_isbn integer not null, primary key(course,sem,book_isbn), foreign key (course) references course (course) on delete cascade, foreign key (book_isbn) references text (book_isbn) on delete cascade );

Table created.

INSERTING TUPLES

SQL>insert into book_adoption values('&course','&sem','&book_isbn');

CM/SB/S Page 20 2012-2013


Enter value for course: 1Enter value for sem: 5Enter value for book_isbn: 111old 2: ('&course','&sem','&book_isbn')new 2: ('001','5','111')

1 row created.

SQL> select * from book_adoption;

COURSE SEM BOOK_ISBN----------------------------------------------------- 1 5 111 1 6 222 1 4 333 3 6 444 4 4 555

QUERIES

1. Demonstrate how you add a new text book to the database and make this book be adopted by some department

SQL> insert into text values ('666','computer networks','venugopal','ferouzon');

1 row created.

SQL> insert into book_adoption values ('5','6','666');

1 row created.RESULT:

SQL> select * from text;

BOOK_ISBN BOOK_TITLE PUBLISHER---------- ------------------------------ ------------------------------AUTHOR------------------------------ 111 linux pearsonsumithaba das

222 compiler design tmhtenenbaum

333 computer graphics pearsonmicheal folk

444 data structures venugopal

CM/SB/S Page 21 2012-2013


tmh

555 file structures godsehimalaya

666 computer networks venugopalferouzon

6 rows selected.

SQL> select * from book_adoption;

COURSE SEM BOOK_ISBN------------------------------------------------- 1 5 111 1 6 222 1 4 333 3 6 444 4 4 555 5 6 666

6 rows selected.

2. Produce a list of textbooks in the alphabetic order for courses offered by the ‘CSE’ department that use more than two books

SQL> select c.course, t.book_isbn, t.book_title from course c, book_adoption ba, text t where c.course=ba.course and c.dept='cse' and ba.book_isbn=t.book_isbn and exists (select count(course) from book_adoption where course=c.course group by course having count(course)>2) order by book_title;

RESULT

COURSE BOOK_ISBN BOOK_TITLE------------------------------------------------------------- 1 222 compiler design 1 333 computer graphics 1 111 linux

3. List any department that has all its adopted books published by a specificpublisher.

SQL> select distinct c.dept from course c,text t,book_adoption ba where c.course=ba.course and t.book_isbn=ba.book_isbn and t.publisher='venugopal'

CM/SB/S Page 22 2012-2013


and t.publisher=all(select t1.publisher from course c1,book_adoption ba1,text t1 where ba1.book_isbn=t1.book_isbn and ba1.course=c1.course and c.dept=c1.dept);

RESULT

DEPT--------------------------ise

CM/SB/S Page 23 2012-2013


Problem-4

4. The following tables are maintained by a book dealer.

AUTHOR (author-id:int, name:string, city:string, country:string)

PUBLISHER (publisher-id:int, name:string, city:string, country:string)

CATALOG (book-id:int, title:string, author-id:int, publisher-id:int, category-

id:int, year:int, price:int)

CATEGORY (category-id:int, description:string)

ORDER-DETAILS (order-no:int, book-id:int, quantity:int)


the foreign keys.


iii. Give the details of the authors who have 2 or more books in the

catalog and the price of the books is greater than the average price of

the books in the catalog and the year of publication is after 2000.

iv. Find the author of the book which has maximum sales.

v. Demonstrate how you increase the price of books published by a

specific publisher by 10%.


vii. Create suitable front end for querying and displaying the results

TABLE CREATION

SQL> create table author ( authid number(4) primary key, name varchar(10), city varchar(10), country varchar(10) );

Table created.

INSERTING TUPLES

SQL> insert into author values ('&authid','&name','&city','&country');Enter value for authid: 11Enter value for name: aaaEnter value for city: delhi

CM/SB/S Page 24 2012-2013


Enter value for country: indiaold 2: ('&authid','&name','&city','&country')new 2: ('11','aaa','delhi','india')

1 row created.

SQL> select * from author;

AUTHID NAME CITY COUNTRY---------------------------------------------------------------------------------- 11 aaa delhi india 12 bbb banglore india 13 ccc luxar africa 14 ddd paris france 15 eee poland germany

TABLE CREATION

SQL> create table publisher ( pubid number(4) primary key, name varchar(10), city varchar(10), country varchar(10) );

Table created.

INSERTING TUPLES

SQL> insert into publisher values ('&pubid','&name','&city','&country');Enter value for pubid: 21Enter value for name: macgrawEnter value for city: delhiEnter value for country: indiaold 2: ('&pubid','&name','&city','&country')new 2: ('21','macgraw','delhi','india')

1 row created.

SQL> select * from publisher;

PUBID NAME CITY COUNTRY--------------------------------------------------------------------------------------------------------- 21 macgraw delhi india 22 subash paris france 23 annai banglore india 24 DSC delhi india 25 KSC poland germany

CM/SB/S Page 25 2012-2013


TABLE CREATION

SQL> create table category ( catid number(4) primary key, des varchar(20) );

Table created.

INSERTING TUPLES

SQL> insert into category values ('&catid','&des');Enter value for catid: 31Enter value for des: sys s/wold 2: ('&catid','&des')new 2: ('31','sys s/w')

1 row created.SQL> insert into category values ('32','os')SQL> insert into category values ('33','dbms')SQL> insert into category values ('34','gt')SQL> insert into category values ('35','oops')

SQL> select * from category;

CATID DES----------------------------------- 31 sys s/w 32 os 33 dbms 34 gt 35 oopsTABLE CREATION

SQL> create table catalog ( bookid number(4) primary key, title varchar(20), authid number(4) references author(authid), pubid number(4) references publisher(pubid), catid number(4) references category(catid), year number(4), price number(4) );

Table created.INSERTING TUPLES

SQL> insert into catalog values ('&bookid','&title','&authid','&pubid','&catid','&year','&price');

CM/SB/S Page 26 2012-2013


Enter value for bookid: 41Enter value for title: logic designEnter value for authid: 11Enter value for pubid: 21Enter value for catid: 31Enter value for year: 2003Enter value for price: 300old 2: ('&bookid','&title','&authid','&pubid','&catid','&year','&price')new 2: ('41','logic design','11','21','31','2003','300')

1 row created.

SQL> select * from catalog;

BOOKID TITLE AUTHID PUBID CATID YEAR PRICE----------------------------------------------------------------------------------------------------------- 41 logic design 11 21 31 2003 300 42 graph theory 11 22 32 2002 480 43 c++ appl 13 23 33 1992 330 44 dbms mang 14 24 34 2008 225 45 c prog 14 25 35 2000 150TABLE CREATION

SQL> create table order_details ( ordno number(4) primary key, bookid number(4) references catalog(bookid), quantity number(4) );

Table created.

INSERTING TUPLES

SQL> insert into order_details values ('&ordno','&bookid','&quantity');Enter value for ordno: 91Enter value for bookid: 41Enter value for quantity: 50old 2: ('&ordno','&bookid','&quantity')new 2: ('91','41','50')

1 row created.

SQL> select * from order_details;

ORDNO BOOKID QUANTITY------------------------------------------------------------------

CM/SB/S Page 27 2012-2013


91 41 50 92 42 100 93 43 150 94 43 25 95 45 160

QUERIES

1. Give the details of the authors who have 2 or more books in the catalog and the price of the books is greater than the average price of the books in thecatalog and the year of publication is after 2000.

SQL> select a.authid, a.name ,a.city, count(*) as count from author a,catalog c where a.authid=c.authid and c.year>2000 and c.price>=(select avg (price) from catalog) group by(a.authid, a.name, city) having count(*)>=2;

RESULT

AI NAME CITY COUNT---------------------------------------------------------------11 aaa delhi 2

2. Find the author of the book which has maximum sales.

SQL> select distinct(a.name) from author a,catalog c,order_details odm where a.authid=c.authid and odm.bookid=c.bookid and exists (select od.bookid,sum(od.quantity) from order_details od where od.bookid=odm.bookid group by bookid having sum(od.quantity)>=all(select sum(quantity) from order_details group by bookid));RESULT

NAME---------- ccc3. Demonstrate how you increase the price of books published by a specificpublisher by 10%.

SQL> update catalog set price = 1.1 * price

CM/SB/S Page 28 2012-2013


where pubid in (select pubid from publisher where name=’DSC’);

RESULT

SQL> select * from catalog;

BOOKID TITLE AUTHID PUBID CATID YEAR PRICE----------------------------------------------------------------------------------------------------------- 41 logic design 11 21 31 2003 300 42 graph theory 11 22 32 2002 480 43 c++ appl 13 23 33 1992 330 44 dbms mang 14 24 34 2008 248 45 c prog 14 25 35 2000 150

CM/SB/S Page 29 2012-2013


PROBLEM-5

5. Consider the following database for a banking enterprise

BRANCH(branch-name:string, branch-city:string, assets:real)

ACCOUNT(accno:int, branch-name:string, balance:real)

DEPOSITOR(customer-name:string, accno:int)

CUSTOMER(customer-name:string, customer-street:string, customer-city:string)

LOAN(loan-number:int, branch-name:string, amount:real)

BORROWER(customer-name:string, loan-number:int)

i. Create the above tables by properly specifying the primary keys

and the foreign keys

ii. Enter at least five tuples for each relation

iii. Find all the customers who have at least two accounts at the Main

branch.

iv. Find all the customers who have an account at all the branches

located in a specific city.

v. Demonstrate how you delete all account tuples at every branch

located in a specific city.


vii. Create suitable front end for querying and displaying the results

TABLE CREATIONSQL> create table branch ( branch_name varchar(25) primary key, branch_city varchar(20) not null, assets decimal(10,2) not null );

Table created.

INSERTING TUPLES

SQL> insert into branch values('jaynagar’,'bangalore','15000000');

BRANCH_NAME BRANCH_CITY ASSETS

CM/SB/S Page 30 2012-2013


------------------------------------------------------------------------------------ jaynagar bangalore 15000000 basavanagudi bangalore 25000000 noida delhi 50000000 marine drive mumbai 40000000 green park delhi 30000000

TABLE CREATION

SQL> create table account ( accno int primary key, branch_name varchar(25) not null, balance decimal(10,2) not null, foreign key(branch_name) references branch(branch_name) );

Table created.

INSERTING TUPLES

SQL>insert into account values(‘123’,’jaynagar’,’25000’);

SQL> select * from account;

ACCNO BRANCH_NAME BALANCE---------- ------------------------- ------------------------------------ 123 jaynagar 25000 156 jaynagar 30000 456 basavanagudi 15000 789 noida 25000 478 marine drive 48000 778 green park 60000 189 basavanagudi 48888

TABLE CREATION

SQL> create table customer ( customer_name varchar(25) primary key, customer_street varchar(25) not null, customer_city varchar(25) not null );

Table created.INSERTING TUPLESSQL> insert into customer values(‘ramu’,’jaynagar’,’bangalore’);

SQL> select * from customer;

CM/SB/S Page 31 2012-2013


CUSTOMER_NAME CUSTOMER_STREET CUSTOMER_CITY-------------------------------------------------------------------------------------------------------ramu jaynagar bangalorekumar basavanagudi bangalorejohn noida delhimike marine drive mumbaisachin green park delhi

TABLE CREATION

SQL> create table depositor ( customer_name varchar(25) not null, accno int not null, foreign key (customer_name) references customer (customer_name) foreign key(accno) references account(accno) on delete cascade );

Table created.

INSERTING TUPLES

SQL> insert into depositor values(‘ramu’,’123’);

SQL> select *from depositor;

CUSTOMER_NAME ACCNO--------------------------------------------------ramu 123ramu 156ramu 189kumar 456john 789mike 478sachin 778TABLE CREATION

SQL> create table loan ( loan_number int primary key, branch_name varchar(25) not null, amount decimal(10,2) not null, foreign key(branch_name) references branch (branch_name) );

Table created.INSERTING TUPLES

SQL> insert into loan values(‘1111’,’jaynagar’,’250000’);

CM/SB/S Page 32 2012-2013


SQL> select *from loan;

LOAN_NUMBER BRANCH_NAME AMOUNT----------------------------------------------------------------------------------- 1111 jaynagar 250000 2222 basavanagudi 350000 3333 noida 150000 4444 marine drive 1500000 5555 green park 7500000

TABLE CREATION

SQL> create table borrower ( customer_name varchar(25) not null, loan_number int not null, foreign key(customer_name) references customer (customer_name), foreign key (loan_number) references loan (loan_number), primary key(customer_name,loan_number) );

Table created.

INSERTING TUPLES

SQL> insert into borrower values(‘ramu’,’1111’);

SQL> select *from borrower;

CUSTOMER_NAME LOAN_NUMBER-------------------------------------------------------------ramu 1111kumar 2222john 3333mike 4444sachin 5555

QUERIES

1. Find all the customers who have atleast two accounts at the Main branch.

SQL> select distinct(customer_name), count(*) from account a, depositor d where a.accno=d.accno and d.accno in (select accno from account where branch_name=’jaynagar’) group by d.customer_name having count(*)>=2;

CM/SB/S Page 33 2012-2013


RESULT

CUSTOMER_NAME COUNT(*)------------------------------------------------------ramu 2

2. Find all the customers who have an account at all the branches located in a specific city.

SQL> select d.customer_name from account a, depositor d, branch b where b.branch_name=a.branch_name and a.accno=d.accno and b.branch_city=’bangalore’ having count(distinct b.branch_name)=(select count(branch_name) from branch where branch_city=’bangalore’) group by customer_name;

RESULT

CUSTOMER_NAME-------------------------ramu

3. Demonstrate how you delete all account tuples at every branch located in a specific city.

SQL> delete from account where branch_name in (select branch_name from branch where branch_city='delhi');2 rows deleted.

RESULT

SQL> select *from account;

ACCNO BRANCH_NAME BALANCE----------------------------------------------------------------------- 123 jaynagar 25000 156 jaynagar 30000 456 basavanagudi 15000 478 marine drive 48000 189 basavanagudi 48888SQL> select *from depositor;

CUSTOMER_NAME ACCNO------------------------------------------------ramu 123

CM/SB/S Page 34 2012-2013


ramu 156ramu 189kumar 456mike 478

CM/SB/S Page 35 2012-2013

Date post:	12-Nov-2014
Category:	Documents
Upload:	chandra-mohan
View:	46 times
Download:	7 times

DBMS_Lab_10CSL57_2012

Documents