Database Management System Laboratory 16MCA28
Database Management System Laboratory 16MCA28
Database Management System Laboratory 16MCA28
Overview of Database
A Database is a collection of related data organized
in a way that data can be easily accessed, managed and updated. Any
piece of information can be a data, for example name of your
school. Database is actually a place where related piece of
information is stored and various operations can be performed on
it.
A DBMS is software that allows creation, definition
and manipulation of database. Dbms is a tool used to perform any
kind of operation on data in database. Dbms also provides
protection and security to database. It maintains data consistency
in case of multiple users. Here are some examples of popular dbms,
MySql, Oracle, Sybase, Microsoft Access and IBM DB2 etc.
SQL is a programming language for Relational Databases. It is
designed over relational algebra and tuple relational calculus. SQL
comes as a package with all major distributions of RDBMS.
SQL comprises both data definition and data manipulation
languages. Using the data definition properties of SQL, one can
design and modify database schema, whereas data manipulation
properties allows SQL to store and retrieve data from database.
Data Definition Language
SQL uses the following set of commands to define database schema
−
CREATE
Creates new databases, tables and views from RDBMS.
For example −
Create database tutorialspoint;
Create table article;
Create view for_students;
DROP
Drops commands, views, tables, and databases from RDBMS.
For example−
Drop object_type object_name;
Drop database tutorialspoint;
Drop table article;
Drop view for_students;
ALTER
Modifies database schema.
Alter object_type object_name parameters;
For example−
Alter table article add subject varchar;
This command adds an attribute in the relation article with the
name subject of string type.
Data Manipulation Language
SQL is equipped with data manipulation language DML. DML
modifies the database instance by inserting, updating and deleting
its data. DML is responsible for all from data modification in a
database. SQL contains the following set of commands in its DML
section −
SELECT/FROM/WHERE
INSERT INTO/VALUES
UPDATE/SET/WHERE
DELETE FROM/WHERE
These basic constructs allow database programmers and users to
enter data and information into
the database and retrieve efficiently using a number of filter
options.
SELECT/FROM/WHERE
SELECT − this is one of the fundamental query command of SQL. It
is similar to the projection operation of relational algebra. It
selects the attributes based on the condition described by WHERE
clause.
FROM − this clause takes a relation name as an argument from
which attributes are to be selected/projected. In case more than
one relation names are given, this clause corresponds to Cartesian
product.
WHERE − this clause defines predicate or conditions, which must
match in order to qualify the attributes to be projected.
For example −
Select author_name
From book_author
Where age > 50;
This command will yield the names of authors from the relation
book_author whose age is greater than 50.
INSERT INTO/VALUES
This command is used for inserting values into the rows of a
table relation.
Syntax−
INSERT INTO table (column1 [, column2, column3 ... ]) VALUES
(value1 [, value2, value3
... ])
Or
INSERT INTO table VALUES (value1, [value2, ... ])
For example −
INSERT INTO tutorialspoint (Author, Subject) VALUES
("anonymous", "computers");
UPDATE/SET/WHERE
This command is used for updating or modifying the values of
columns in a table relation.
Syntax −
UPDATE table_name SET column_name = value [, column_name = value
...] [WHERE condition]
For example −
UPDATE tutorialspoint SET Author="webmaster" WHERE
Author="anonymous";
DELETE/FROM/WHERE
This command is used for removing one or more rows from a table
relation.
Syntax −
DELETE FROM table_name [WHERE condition];
For example −
DELETE FROM tutorialspoints
WHERE Author="unknown";
1) Create the following tables with properly specifying Primary
keys, foreign keys and solve the following queries.
i. BRANCH(Branchid, Branchname, HOD)
ii. STUDENT(USN, Name, Address, Branchid, sem)
iii. BOOK(Bookid, Bookname, Authorid, Publisher, Branchid)
iv. AUTHOR(Authorid, Authorname, Country, age)
v. BORROW(USN ,Bookid, Borrowed_Date)
Queries:
1 List the details of Students who are all studying in 2ndsem
MCA.
2 List the students who are not borrowed any books.
3 Display the USN, Student name, Branch_name, Book_name,
Author_name, Books_Borrowed_Date of 2nd sem MCA Students who
borrowed books.
4 Display the number of books written by each Author.
5 Display the student details who borrowed more than two
books.
6 Display the student details who borrowed books of more than
one Author.
7 Display the Book names in descending order of their names.
8 List the details of students who borrowed the books which are
all published by the same Publisher.
CREATE TABLE BRANCH
(
BRANCHID NUMBER PRIMARY KEY,
BRANCHNAME VARCHAR(20),
HOD VARCHAR(20)
);
INSERT INTO BRANCH VALUES (1,'MCA','SRINIVAS RAO');
INSERT INTO BRANCH VALUES (2,'BE','TIPPESWAMY');
INSERT INTO BRANCH VALUES (3,'BE_EC','RAVISHANKAR');
INSERT INTO BRANCH VALUES (4,'BE_ISE','JOHN');
CREATE TABLE STUDENT
(
USN VARCHAR(30) PRIMARY KEY,
NAME VARCHAR(40),
ADDRESS VARCHAR(60),
BRANCHID NUMBER REFERENCES BRANCH,
SEM NUMBER
);
INSERT INTO STUDENT VALUES ( '1ST16MCA01',
'ADARSH','KERALA',1,2);
INSERT INTO STUDENT VALUES ( '1ST16CS02',
'ANIS','KERALA',2,2);
INSERT INTO STUDENT VALUES ( '1ST16EC03',
'ARCHANA','CHANDIGARH',3,3);
INSERT INTO STUDENT VALUES ( '1ST16IS01', 'NAYON','WEST
BENGAL',4,4);
INSERT INTO STUDENT VALUES ( '1ST16MCA02',
'RAHUK','BIHAR',1,2);
CREATE TABLE AUTHOR
(
AUTHORID NUMBER PRIMARY KEY,
AUTHORNAME VARCHAR(30),
COUNTRY VARCHAR(30),
AGE NUMBER
);
INSERT INTO AUTHOR VALUES (11,'WILLIAM
STALLING','AMERICA',30);
INSERT INTO AUTHOR VALUES (12,'PETERSON','CANADA', 40);
INSERT INTO AUTHOR VALUES (13,'PADMAREDDY','INDIA',35);
INSERT INTO AUTHOR VALUES (14,'NAMRATHA','INDIA',27);
CREATE TABLE BOOK
(
BOOKID NUMBER PRIMARY KEY,
BOOKNAME VARCHAR(30),
AUTHORID NUMBER REFERENCES AUTHOR,
PUBLISHER VARCHAR(30),
BRANCHID NUMBER REFERENCES BRANCH
);
INSERT INTO BOOK VALUES (111,'MCADBMS',11,'WILLIAM
PUBLISHER',1);
INSERT INTO BOOK VALUES (222,'CSDBMS',12,'PETERSON
PUBLISHER',2);
INSERT INTO BOOK VALUES (333,'ISEDBMS',13,'PADMAREDDY
PUBLISHER',3);
INSERT INTO BOOK VALUES (444,'ECDBMS',14,'NAMRATHA
PUBLISHER',4);
CREATE TABLE BORROW
(
USN VARCHAR(30) REFERENCES STUDENT,
BOOKID NUMBER REFERENCES BOOK,
BORROWED_DATE DATE
);
INSERT INTO BORROW VALUES( '1ST16MCA01',111,date
'2010-11-02');
INSERT INTO BORROW VALUES( '1ST16CS02',222,date
'2014-1-02');
INSERT INTO BORROW VALUES( '1ST16IS01',333,date
'2014-4-05');
INSERT INTO BORROW VALUES( '1ST16IS01',333,date
'2015-5-07');
INSERT INTO BORROW VALUES( '1ST16IS01',333,date
'2017-4-05');
1. List the details of Students who are all studying in 2ndsem
MCA.
SQL> SELECT * FROM STUDENT WHERE SEM=2;
USN
NAME
ADDRESS
BRANCHID
SEM
1ST16MCA01
ADARSH
KERALA
1
2
1ST16CS02
ANIS
KERALA
2
2
1ST16MCA02
RAHUK
BIHAR
1
2
2) List the students who are not borrowed any books.
SQL> SELECT * FROM STUDENT WHERE USN NOT IN ( SELECT USN FROM
BORROW);
USN
NAME
ADDRESS
BRANCHID
SEM
1ST16EC03
ARCHANA
CHANDIGARH
3
3
1ST16MCA02
RAHUK
BIHAR
1
2
3) Display the USN, Student name, Branch_name, Book_name,
Author_name, Books_Borrowed_Date of 2nd sem MCA Students who
borrowed books.\
select distinct
a1.USN,a1.NAME,c.BOOKNAME,d.AUTHORNAME,a2.BORROWED_DATE from
STUDENT a1 join branch br on a1.BRANCHID=br.BRANCHID join BORROW a2
on a1.USN=a2.usn join book c on a2.BOOKID=c.BOOKID join author d on
c.AUTHORID=d.AUTHORID where a1.SEM=2 and br.BRANCHNAME='MCA';
USN NAME BOOKNAME AUTHORNAME BORROWED_
1ST16MCA01 ADARSH MCADBMS WILLIAM STALLING 02-NOV-10
4) Display the number of books written by each Author
SQL> SELECT AUTHORNAME, COUNT(*) FROM AUTHOR, BOOK WHERE
AUTHOR.AUTHORID=BOOK.A
UTHORID GROUP BY AUTHOR.AUTHORNAME;
AUTHORNAME COUNT(*)
------------------------------ ----------
PETERSON 1
WILLIAM STALLING 1
PADMAREDDY 1
NAMRATHA 1
5) Display the student details who borrowed more than two
books.
select * from STUDENT where USN in (select USN from BORROW group
by USN having COUNT(*)>2);
select * from STUDENT where USN in (select USN from BORROW group
by USN hav
ing COUNT(*)>2);
USN NAME ADDRESS BRANCHID SEM
1ST16IS01 NAYON WEST BENGAL 4 4
6) Display the student details who borrowed books of more than
one Author.
select * from student where usn in (select usn from borrow a
join book b on a.BOOKID=b.BOOKID group by usn, b.AUTHORID having
count(b.AUTHORID)>1);
USN NAME ADDRESS BRANCHID SEM
1ST16IS01 NAYON WEST BENGAL 4 4
7) Display the Book names in descending order of their
names.
SQL> SELECT BOOKNAME FROM BOOK ORDER BY BOOKNAME DESC;
BOOKNAME
------------------------------
MCADBMS
ISEDBMS
ECDBMS
CSDBMS
8 List the details of students who borrowed the books which are
all published by the same Publisher.
select * from student where usn in(select a.USN from student a
join borrow b on b.USN=a.USN join
book c on b.BOOKID=c.BOOKID group by a.USN having
count(c.PUBLISHER)=1);
USN
NAME
ADDRESS
BRANCHID
SEM
1ST16MCA01
ANISH
49/5/1C
1
2
1ST16MCA03
DIPAK
49/8/1C
3
3
E-R DIAGRAM :
RELATION DIAGRAM :
2) Design an ER-diagram for the following scenario, Convert the
same into a relational model and then solve the following
queries.
Consider a Cricket Tournament “ABC CUP” organized by an
organization. In the tournament there are many teams are contesting
each having a Teamid,Team_Name, City, a coach. Each team is
uniquely identified by using Teamid. A team can have many Players
and a captain. Each player is uniquely identified by Playerid,
having a Name, and multiple phone numbers,age. A player represents
only one team. There are many Stadiums to conduct matches. Each
stadium is identified using Stadiumid, having a
stadium_name,Address ( involves city,area_name,pincode).A team can
play many matches. Each match played between the two teams in the
scheduled date and time in the predefined Stadium. Each match is
identified uniquely by using Matchid. Each match won by any of the
one team that also wants to record in the database. For each match
man_of_the match award given to a player.
Queries:
1 Display the youngest player (in terms of age) Name, Team name
, age in which he belongs of the tournament.
2 List the details of the stadium where the maximum number of
matches were played.
3 List the details of the player who is not a captain but got
the man_of _match award at least in two matches.
4 Display the Team details who won the maximum matches.
5 Display the team name where all its won matches played in the
same stadium.
SQL> connect
Enter user-name: SYSTEM
Enter password:
Connected.
SQL> create table team ( tid int primary key, tname
varchar(20), coach varchar
(20), captain_pid int, city varchar(20));
Table created.
SQL> create table player ( pid int primary key, pname
varchar(2), age int, tid int references team(tid));
Table created.
SQL> create table stadium (sid int primary key, sname
varchar(20), picode number(8), city varchar(20), area
varchar(20));
Table created.
SQL> create table match (mid int primary key, mdate date,
time varchar(6), sid int references stadium(sid), team1_id int
references team(tid), team2_id int references team(tid),
winning_team_id int references team(tid), man_of_match int
references player(pid));
Table created.
SQL> create table player_phone ( pid int references
player(pid), phone int , primary key(pid,phone));
Table created.
insert into team values(1,'India','Virat',11,'Delhi');
insert into team values(2,'Australia','Smith',22,'Sydney');
insert into team
values(3,'WestIndies','Sammy',33,'Kingstown');
insert into team
values(4,'Srilanka','Sangakara',44,'Colombo');
select * from team;
TID TNAME COACH CAPTAIN_PID CITY
1 India Virat 11 Delhi
2 Australia Smith 22 Sydney
3 WestIndies Sammy 33 Kingstown
4 Srilanka Sangakara 44 Colombo
insert into player values(101,'MS',35,1);
insert into player values(102,'ps',40,2);
insert into player values(103,'MM',26,3);
insert into player values(104,'NO',36,4);
select * from player;
PID PN AGE TID
---------- -- ---------- ----------
101 MS 35 1
102 ps 40 2
103 MM 26 3
104 NO 36 4
insert into stadium
values(201,'chinnaswamy',560010,'bangalore','karnataka');
insert into stadium
values(202,'Eadengarden',200010,'Kolkata','Westbengal');
insert into stadium
values(203,'Melbourne',122010,'Melbourne','Australia');
SQL> select * from stadium;
SID SNAME PICODE CITY AREA
201 chinnaswamy 560010 bangalore karnataka
202 Eadengarden 200010 Kolkata Westbengal
203 Melbourne 122010 Melbourne Australia
insert into match
values(301,date'2010-5-10','10am',201,1,2,1,101);
insert into match
values(302,date'2016-3-15','11am',202,2,3,3,103);
select * from match;
MID MDATE TIME SID TEAM1_ID TEAM2_ID WINNING_TEAM_ID
MAN_OF_MATCH
301 10-MAY-10 10am 201 1 2 1 101
302 15-MAR-16 11am 202 2 3 3
103
insert into player_phone values(101,100000);
insert into player_phone values(102,2100000);
insert into player_phone values(103,23330000);
1) Display the youngest player (in terms of age) Name, Team name
, age in which he belongs of the tournament.
SELECT B.NAME, A.TEAMNAME, B.AGE FROM TEAM A , PLAYER B WHERE
A.TEAMID = B.TEAMID AND B.AGE=(SELECT MIN(AGE) FROM PLAYER ;
2) List the details of the stadium where the maximum number of
matches were played.
SQL> select * from stadium where sid in (select sid from
match group by sid having count(sid) = (select max(count(sid)) from
match group by sid)) ;
SID SNAME PICODE CITY
---------- -------------------- ----------
--------------------
AREA
--------------------
201 chinnaswamy 560010 bangalore
karnataka
202 Eadengarden 200010 Kolkata
Westbengal
3) List the details of the player who is not a captain but got
the man_of _match award at least in two matches.
SQL> select * from player where pid not in ( select
captain_pid from team where
captain_pid not in (select man_of_match from match group by
man_of_match having
count(man_of_match)=2));
PID PN AGE TID
---------- -- ---------- ----------
101 MS 35 1
102 ps 40 2
103 MM 26 3
104 NO 36 4
4) Display the Team details who won the maximum matches.
SQL> select * from team where tid in
2 (select winning_team_id from match group by
winning_team_id
3 having count(winning_team_id)=
4 (select max(count(winning_team_id))
5 from match group by winning_team_id))
6 ;
TID TNAME COACH CAPTAIN_PID
---------- -------------------- --------------------
-----------
CITY
--------------------
1 India Virat 11
Delhi
3 WestIndies Sammy 33
Kingstown
5)
Display the team name where all its won matches played in the
same stadium.
SQL> SELECT MAX(winning_team_id)FROM ( SELECT DISTINCT
winning_team_id, sid
FROM match )GROUP BY sid HAVING COUNT(*) = 1;
MAX(WINNING_TEAM_ID)
--------------------
1
3
3) Consider the following Scenario and design an ER-Diagram, map
the designed ER-diagram into a Relational model.
Consider an organization “ABC” having many employees. An
employee works for one department. Each employee identified by
using Empid, having Name, address ( described as House_no, city,
district, state, pin code) and more than one phone numbers.
Department identified by using Dno, having Dname, Dlocation. Each
Department having a manager . Each department having many
employees. There are many Projects , each project is controlled by
the department. Each Project uniquely identified by Pno, having
Project_name,Project_location. An employee works on many Projects.
Number of hours per week worked on each project by an Employee also
needs to be recorded in the database . A project is worked by many
employees. Each employee supervised by the supervisor. Employee
having many dependents. Dependents having the dependent_name,
gender, age, address. Dependents are identified by Empid.
T1(Empid, Emp_Name,city, district, state, pin_code, phoneno,
Dno,Dname,Dlocation, Dept_mgr_id, Pno, Project_name,
Project_location, Number_of_Hours,Supervisor_Empid, Dependent_name,
gender, address)
Queries:
1. Display the details of the employees who are working on both
the projects having project_no 5 and 10.
2. Display the details of employees having atleast two
dependents.
3. Display the project name on which more number of employees
are working.
4. Retrieve the employees who do not have any dependents.
5. Display the Employee details whose total number of hours per
week working on various projects is maximum than all other
employees.
6. create a view to display the number of employees working in
each department.
CREATE TABLE EMPLOYEE (EMPID INTEGER PRIMARY KEY, EMPNAME
VARCHAR(20), HOUSENUM NUMBER, CITY VARCHAR(20),DISTRICT
VARCHAR(20),STATE VARCHAR(20),PINCODE NUMBER, PHONENUM1 NUMBER,
PHONENUM2 NUMBER);
DESC EMPLOYEE;
Name Null? Type
----------------------------------------- --------
----------------------------
EMPID NOT NULL NUMBER(38)
EMPNAME VARCHAR2(20)
HOUSENUM NUMBER
CITY VARCHAR2(20)
DISTRICT VARCHAR2(20)
STATE VARCHAR2(20)
PINCODE NUMBER
PHONENUM1 NUMBER
PHONENUM2 NUMBER
INSERT INTO EMPLOYEE VALUES
(1,'NAMRATHA',200,'DHARWAD','DHARWAD','KARNATAKA',560010,9632223445,9474344434);
INSERT INTO EMPLOYEE VALUES
(2,'SNEHA',300,'BANGALORE','BANGALORE','KARNATAKA',580070,9635273845,9478328434);
INSERT INTO EMPLOYEE VALUES
(3,'NETRA',400,'BELGAUM','BELGAUM','KARNATAKA',540030,9335273945,9276380434);
INSERT INTO EMPLOYEE VALUES
(4,'MANJU',500,'HYDERABAD','HYDERABAD','ANDRAPRADESH',340034,9335263845,726684444);
INSERT INTO EMPLOYEE VALUES
(5,'ADARSH',600,'ALPHE','TRISSUR','KERALA',340038,9442334855,5266844433);
INSERT INTO EMPLOYEE VALUES
(6,'ABC',600,'RAJAJINA','BELGAUM','KARNATAKA',540030,9335273945,9276380434);
INSERT INTO EMPLOYEE
VALUES(7,'DEFC',700,'ABCDEFGH','BELGAUM','KARNATAKA',543334,53463,56277676);
SQL> CREATE TABLE DEPARTMENT (DEPTNO NUMBER PRIMARY KEY,
DEPTNAME VARCHAR(20),DEPTLOCATION VARCHAR(20), DEPTMANAGERID
REFERENCES EMPLOYEE(EMPID));
DESC DEPARTMENT;
Name Null? Type
----------------------------------------- --------
----------------------------
DEPTNO NOT NULL NUMBER
DEPTNAME VARCHAR2(20)
DEPTLOCATION VARCHAR2(20)
DEPTMANAGERID NUMBER
INSERT INTO DEPARTMENT VALUES(11,'DEVELOPER','BANGALORE',1);
INSERT INTO DEPARTMENT VALUES(12,'TESTER','TRISSUR',5);
INSERT INTO DEPARTMENT
VALUES(13,'NETWORKENGINEER','CHENNAI',2);
INSERT INTO DEPARTMENT
VALUES(14,'MAINTAINANCE','HYDERABAD',3);
INSERT INTO DEPARTMENT VALUES(15,'SECURITY','BANGALORE',4);
SELECT * FROM DEPARTMENT;
DEPTNO DEPTNAME DEPTLOCATION DEPTMANAGERID
---------- -------------------- --------------------
-------------
11 DEVELOPER BANGALORE 1
12 TESTER TRISSUR 5
13 NETWORKENGINEER CHENNAI 2
14 MAINTAINANCE HYDERABAD 3
15 SECURITY BANGALORE 4
CREATE TABLE PROJECT (PNO NUMBER PRIMARY KEY, PNAME
VARCHAR(20),PLOCATION VARCHAR(20),NUMBEROFHOURS NUMBER,
SUPERVISOREMPID REFERENCES EMPLOYEE(EMPID));
SQL> DESC PROJECT;
Name Null? Type
----------------------------------------- --------
----------------------------
PNO NOT NULL NUMBER
PNAME VARCHAR2(20)
PLOCATION VARCHAR2(20)
NUMBEROFHOURS NUMBER
SUPERVISOREMPID NUMBER
SQL> INSERT INTO PROJECT VALUES(5,'ANDROID
APPLICATION','BANGALORE',20,1);
1 row created.
SQL> INSERT INTO PROJECT VALUES(6,'MOBILE
APPLICATION','HYDERABAD',30,2);
1 row created.
SQL> INSERT INTO PROJECT
VALUES(7,'WEBSITE','CHENNAI',40,3);
1 row created.
SQL> INSERT INTO PROJECT
VALUES(8,'SMP','BANGALORE',50,4);
1 row created.
SQL> INSERT INTO PROJECT VALUES(10,'WIRELESS
COMMU','BANGALORE',60,5);
1 row created.
SQL> CREATE TABLE DEPENDENT (DEPENDENTNAME VARCHAR(20),
GENDER VARCHAR(20), ADDRESS VARCHAR(20), AGE NUMBER, DEPENDENTEMPID
REFERENCES EMPLOYEE(EMPID));
Table created.
SQL> DESC DEPENDENT;
Name Null? Type
----------------------------------------- --------
----------------------------
DEPENDENTNAME VARCHAR2(20)
GENDER VARCHAR2(20)
ADDRESS VARCHAR2(20)
AGE NUMBER
DEPENDENTEMPID NUMBER
SQL>
SQL> INSERT INTO DEPENDENT VALUES(
'MANJUNATH','MALE','DHARWAD',30,1);
1 row created.
SQL> INSERT INTO DEPENDENT VALUES(
'SANTOSH','MALE','BANGALORE',28,2);
1 row created.
SQL> INSERT INTO DEPENDENT VALUES(
'VIJAY','MALE','BELGAUM',34,3);
1 row created.
SQL> INSERT INTO DEPENDENT VALUES(
'AKSHATHA','FEMALE','TRISSUR',24,4);
1 row created.
SQL> INSERT INTO DEPENDENT VALUES(
'ANU','FEMALE','HYDERABAD',36,5);
1 row created.
SQL> INSERT INTO DEPENDENT VALUES(
'SANVI','FEMALE','DHARWAD',6,1);
1 row created.
SQL> SELECT * FROM DEPENDENT;
DEPENDENTNAME GENDER ADDRESS AGE
-------------------- -------------------- --------------------
----------
DEPENDENTEMPID
--------------
MANJUNATH MALE DHARWAD 30
1
SANTOSH MALE BANGALORE 28
2
VIJAY MALE BELGAUM 34
3
DEPENDENTNAME GENDER ADDRESS AGE
-------------------- -------------------- --------------------
----------
DEPENDENTEMPID
--------------
AKSHATHA FEMALE TRISSUR 24
4
ANU FEMALE HYDERABAD 36
5
SANVI FEMALE DHARWAD 6
1
QUERIES:
2) Display the details of employees having at least two
dependents
SQL> SELECT * FROM EMPLOYEE WHERE( SELECT COUNT(*) FROM
DEPENDENT WHERE DEPENDENTEMPID=EMPID)>=2;
EMPID ENAME HOUSENO CITY DISTRICT STATE PINCODE PHONENUM1
PHONENUM2
1 NAMRATHA 200 DHARWAD DHARWAD KARNATAKA 560010 9632223445
9474344434
3) Display the project name on which more number of employees
are working
SQL> SELECT EMPID, PNO FROM EMPLOYEE INNER JOIN PROJECT ON
EMPLOYEE.EMPID =PRO
JECT.SUPERVISOREMPID WHERE EMPLOYEE.EMPID IN (SELECT EMPID FROM
PROJECT GROUP BY
EMPID HAVING COUNT (EMPID)>2);
EMPID PNO
---------- ----------
1 5
2 6
3 7
4 8
5 10
5 11
4) Retrieve the employees who do not have any dependents.
SQL> SELECT * FROM EMPLOYEE WHERE NOT EXISTS (SELECT * FROM
DEPENDENT WHERE EMPI
D=DEPENDENTEMPID);
EMPID ENAME HOUSENO CITY
---------- -------------------- ----------
--------------------
DISTRICT STATE PINCODE PHONENUM1 PHONENUM2
-------------------- -------------------- ---------- ----------
----------
7 DEFC 700 ABCDEFGH
BELGAUM KARNATAKA 543334 53463 56277676
Prepared By: Namratha K, Asst. Prof. Sambhram Institute of
Technology, Bangalore Page 1
Prepared By: Namratha K, Asst. Prof. Sambhram Institute of
Technology, BangalorePage 24
