Q1.Create following two tables including integrity constraints using SQL Command: Table Name : EmployeeField Name Data Type Field Size ConstraintsEmployee ID Number 8 Primary KeyEmployee Name Text 20 Not NullStreet Text 15City Text 3 ‘Del’,’Bom’,’Cal’

Table Name : WorksField Name Data Type Field Size ConstraintsEmployee ID Number 8 Foreign Key,

Primary KeyCompany Name Text 20 Primary KeySalary Number 10,2 >10,000 and < 25,000

Perform the following operations using SQL Commands

(i) List all the employees names working in ‘FBC’(ii) Find the number of employees earning a salary > average salary of all employees.

Ans:- create table employee(employeeid number(8) PRIMARY KEY,employeename varchar2(20) NOT NULL,street varchar2(15),city varchar2(3) CHECK (city IN ( 'del','bom','cal')));

Output:-

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select * from employee;

Output:-

create table works( empid number(8) , FOREIGN KEY(empid) REFERENCES employee(employeeid),companyname varchar2(20),salary number(10,2) CHECK (10000<SALARY AND SALARY <250000));

Output:-

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alter table works add constraint my18primarykey PRIMARY KEY(empid,companyname);

Output:-

select * from works;

Output:-

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select employeename from employee,workswhere companyname='FBC' and employeeid=empid;

Output:-

select count(employeename) from employee,works where salary > (select AVG(salary) from works) and employeeid=empid ;

Output:-

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Q2.Create following tables :Student (StudentID, Student_name)Registered (StudentID, CoursedID, dt_of_Join)

Add the following constraints to the table using alter command:(i) Primary Key and respective foreign keys for all tables.(ii) age > 20 and < 60

List the student id and name of each student along with the total number of courses that the student is registered for.

Ans:-create table student(studentid number(8),stdname varchar2(20));

Output:-

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create table registered(studid number(8) ,courseid number(8),date_of_join date);

Output:-

alter table student add constraint myprimarykey PRIMARY KEY(studentid);

Output:-

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alter table registered add constraint pk_primarykey PRIMARY KEY(studid,courseid);

Output:-

alter table registered add constraint fk_studid FOREIGN KEY(studid) REFERENCES student(studentid);

Output:-

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alter table student add age number(8);

Output:-

alter table student ADD CONSTRAINT check_age CHECK ( 20 < AGE AND AGE < 60);

Output:-

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select studentid,stdname,courseid from student,registered where studentid=studid;

Output:-

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Q3. Create following two tables including integrity constraints using SQL Command:

Table Name : PartField Name Data Type Field Size ConstraintsPart id Text 8 Primary KeyName Text 20 Not Null

Table Name : SubpartField Name Data Type Field Size ConstraintsPart id Text 8 Foreign Key,

Primary KeySub Part id Text 20 Primary KeyCount Number 5 >= 0 and < 100 Cost Number 10,2 Default 0

Perform the following operations using SQL Commands

(i) List all the name of all the subparts of the part id “P-100”.(ii) Find the total cost of part “P-100” including subparts.

Ans:- create table part(partid varchar2(8) primary key,name varchar2(20) not null);

Output:-

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create table subpart(partid varchar2(8),subpartid varchar2(20) primary key,count number(5) check(count between 0 and 100),cost number(10,2) default(0),foreign key(partid) references part);

Output:-

select * from part;

Output:-

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Select * from subpart;

Output:-

select * from subpart where partid='P001';

Output:-

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select sum(cost) from subpart where partid='P001';

Output:-

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Q4.Create following two tables including integrity constraints using SQL Command:

Table Name : StudentField Name Data Type Field Size ConstraintsRoll No Number 8 Primary KeyName Text 20 Not NullAddress Text 50

Table Name : EnrollmentField Name Data Type Field Size ConstraintsRoll No Number 8 Foreign Key,

Primary KeyCourse Text 20 Primary KeyGrade Text 1 ‘A’,’B’,’C’,’D’,’E’

Perform the following operations using SQL Commands

(i) List all the Students enrolled for ‘BBA’ and have grade ‘A’(ii) Find the number of students enrolled in each course.

Ans:-create table student2(rollno number(8) ,name varchar2(20) NOT NULL,address varchar2(50));

Output:-

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create table enrollment(rollnumber number(8) ,course varchar2(20),grade varchar2(1) CHECK(grade IN('A','B','C','D','E')));

Output:-

alter table student2 add constraint pk1_primarykey PRIMARY KEY(rollno);

Output:-

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alter table enrollment add constraint pk2_primarykey PRIMARY KEY(rollnumber,course);

Output:-

alter table enrollment add constraint fk1_rollnumber FOREIGN KEY(rollnumber) REFERENCES student2(rollno);

Output:-

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select * from student2,enrollment where course='bba' and grade='A' and rollno=rollnumber;

Output:-

select count(course) from enrollment GROUP by course;

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Q5.Create following tables :Employee (empno, name, office, age)Books (isbn, title, author, publisher)Loan (empno,isbn,date)

Add the following constraints to the table using alter command:(i) Primary Key and respective foreign key for all tables.(ii) age > 20 and < 60

List names of employees having any book published by “TMH”

Ans:-create table employee2(empno number(8),name varchar2(20),office varchar2(20),age number(8));

Output:-

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create table books(isbn number(20),title varchar2(20),author varchar2(40),publisher varchar2(18));

Output:-

create table loan(employee__no number(20),isbn1 number(20),loandate date);

Output:-

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alter table employee2 add constraint pk8_primarykey PRIMARY KEY(empno);

Output:-

alter table books add constraint pk4_primarykey PRIMARY KEY(isbn);

Output:-

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alter table loan add constraint pk6_primarykey PRIMARY KEY(employee__no,isbn1);

Output:-

alter table loan add constraint fk2_isbn1 FOREIGN KEY(isbn1) REFERENCES books(isbn);

Output:-

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alter table employee2 add CHECK(20 < AGE AND AGE < 60);

Output:-

select name from employee2,books,loan where publisher='TMH' and empno=employee__no and isbn=isbn1;

Output:-

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Q6. Create following two tables including integrity constraints using SQL Command:

Table Name : StudentField Name Data Type Field Size ConstraintsCSTID Number 4 Primary KeyCSTNAME Text 25 Not NullCSTCITY Text 3 Must be-

UDP,MNG,BNG,PJM,MARCSTDEP Number 8,2

Perform the following operations using SQL Commands

(i) Provide constraint names while creating table(ii) Drop the check constraint on CSTCITY(iii) Create a new constraint chkdep to check that the deposit is within the range 1500

to 300000.

Ans:-create table student4(cstid number(4) PRIMARY KEY,cstname varchar2(25) NOT NULL,cstcity varchar2(3) CONSTRAINT check_student CHECK(cstcity IN('UDP','MNG','BNG','PJM','MAR')),cstdep number(8,2));

Output:-

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alter table student4drop constraint check_student;

Output;-

alter table student4 add constraint chkdep CHECK(1500<cstdep and cstdep<300000);

Output;-

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