"Comprehensive Guide to Database Management Systems: Optimizing Data Architecture, Administration, and Performance for Enhanced Organizational Efficiency and Decision-Making"

Experiment 1
AIM: To write a program for SQL Data Definition Language Commands on
sample exercise
Program:
SQL> connect
Enter user-name: system
Enter password: admin
Connected.
SQL> create table emp(id number(10),name varchar(10));
Table created.
SQL> desc emp;
Name Null? Type
----------------------------------------- -------- ----------------------------
ID NUMBER(10)
NAME VARCHAR2(10)
SQL> alter table emp add(dept varchar(10));
Table altered.
SQL> desc emp;
Name Null? Type
----------------------------------------- -------- ----------------------------
ID NUMBER(10)
NAME VARCHAR2(10)
DEPT VARCHAR2(10)
SQL> alter table emp modify dept varchar(20);
Table altered.

SQL> desc emp;
Name Null? Type
----------------------------------------- -------- ----------------------------
ID NUMBER(10)
NAME VARCHAR2(10)
DEPT VARCHAR2(20)
SQL> alter table emp drop column dept;
Table altered.
SQL> desc emp;
Name Null? Type
----------------------------------------- -------- ----------------------------
ID NUMBER(10)
NAME VARCHAR2(10)
SQL> alter table emp rename to emp1;
Table altered.
SQL> desc emp1;
Name Null? Type
----------------------------------------- -------- ----------------------------
ID NUMBER(10)
NAME VARCHAR2(10)
SQL> desc emp2;
Name Null? Type
----------------------------------------- -------- ----------------------------
ID NUMBER(10)
NAME VARCHAR2(10)
DEPT VARCHAR2(10)
SQL> drop table emp2;

Table dropped.
SQL> select * from emp2;
select * from emp2
*
ERROR at line 1:
ORA-00942: table or view does not exist
ID NAME DEPT
---------- ---------- ----------
1 aaa cse
2 aaa cse
3 aaa ece
4 aaa cse
5 aaa cse
SQL> truncate table emp1;
Table truncated.
no rows selected
SQL> desc emp1;
Name Null? Type
----------------------------------------- -------- ----------------------------
ID NUMBER(10)
NAME VARCHAR2(10)
DEPT VARCHAR2(10)
SQL> drop table emp1;
Table dropped.

select * from emp1
*
ERROR at line 1:
SQL> desc emp1;
ERROR:
ORA-04043: object emp1 does not exist
Output:
Result: Hence, the above query has been implemented successfully.

Experiment 2
AIM: To write a program for SQL Data Manipulation Language Commands on
sample exercise
Program:
no rows selected
SQL> alter table emp1 add(dept varchar(10));
Table altered.
SQL> desc emp1;
Name Null? Type
----------------------------------------- -------- ----------------------------
ID NUMBER(10)
NAME VARCHAR2(10)
DEPT VARCHAR2(10)
SQL> insert into emp1(id,name,dept) values (1,'aaa','cse');
1 row created.
ID NAME DEPT
---------- ---------- ----------
1 aaa cse
SQL> insert into emp1 values (2,'bbb','cse');
1 row created.
ID NAME DEPT

---------- ---------- ----------
1 aaa cse
2 bbb cse
SQL> insert into emp1 values(&id,'&name','&dept');
Enter value for id: 3
Enter value for name: ccc
Enter value for dept: cse
old 1: insert into emp1 values(&id,'&name','&dept')
new 1: insert into emp1 values(3,'ccc','cse')
1 row created.
ID NAME DEPT
---------- ---------- ----------
1 aaa cse
2 bbb cse
3 ccc cse
SQL> insert into emp1 values(&id,'&name','&dept');
Enter value for name: ddd
new 1: insert into emp1 values(4,'ddd','cse')
1 row created.
SQL> /
Enter value for name: eee
new 1: insert into emp1 values(5,'eee','cse')

1 row created.
ID NAME DEPT
---------- ---------- ----------
1 aaa cse
2 bbb cse
3 ccc cse
4 ddd cse
5 eee cse
SQL> update emp1 set name='BBB' where id=2;
1 row updated.
ID NAME DEPT
---------- ---------- ----------
1 aaa cse
2 BBB cse
3 ccc cse
4 ddd cse
5 eee cse
SQL> update emp1 set name='CCC',dept='ece' where id=3;
1 row updated.
ID NAME DEPT
---------- ---------- ----------
1 aaa cse
2 BBB cse
3 CCC ece
4 ddd cse
5 eee cse

SQL> update emp1 set name='aaa';
5 rows updated.
ID NAME DEPT
---------- ---------- ----------
1 aaa cse
2 aaa cse
3 aaa ece
4 aaa cse
5 aaa cse
SQL> create table emp2(id number(10),name varchar(10),dept varchar(10));
Table created.
SQL> insert into emp2 select * from emp1;
5 rows created.
ID NAME DEPT
---------- ---------- ----------
1 aaa cse
2 aaa cse
3 aaa ece
4 aaa cse
5 aaa cse
SQL> delete from emp2 where id=1;
1 row deleted.
ID NAME DEPT

---------- ---------- ----------
2 aaa cse
3 aaa ece
4 aaa cse
5 aaa cse
SQL> delete from emp2 where dept='cse';
3 rows deleted.
ID NAME DEPT
---------- ---------- ----------
3 aaa ece
SQL> delete from emp2;
1 row deleted.
no rows selected
no rows selected
Output:

Experiment 3
AIM:To write a program on SQL Data Control Language Commands and
Transaction control commands to the sample exercises
Program:
SQL> connect
Enter password:
Connected.
SQL> create table class1(id number(10),name varchar(10));
Table created.
SQL> insert into class1 values(5,'rahul');
1 row created.
SQL> update class1 set name='raj' where id=5;
1 row updated.
SQL> savepoint A;
Savepoint created.
SQL> insert into class1 values(6,'ram');
1 row created.
SQL> insert into class1 values(7,'vibhav');
1 row created.
SQL> savepoint B;
Savepoint created.

SQL> insert into class1 values(8,'sai');
1 row created.
SQL> savepoint C;
Savepoint created.
SQL> select * from class1;
ID NAME
---------- ----------
5 raj
6 ram
7 vibhav
8 sai
SQL> rollback to B;
Rollback complete.
ID NAME
---------- ----------
5 raj
6 ram
7 vibhav
SQL> rollback to A;
Rollback complete.
ID NAME
---------- ----------

5 raj
SQL> insert into class1 values(6,'ram');
1 row created.
SQL> insert into class1 values(7,'vibhav');
1 row created.
SQL> insert into class1 values(8,'sai');
1 row created.
SQL>savepoint D;
Savepoint created.
SQL> insert into class1 values(9,'siva');
1 row created.
SQL> commit;
Commit complete.
ID NAME
---------- ----------
5 raj
6 ram
7 vibhav
8 sai
9 siva
SQL> rollback to D;

rollback to D
*
ERROR at line 1:
ORA-01086: savepoint 'D' never established
SQL> insert into class1 values(10,'tom');
1 row created.
SQL> savepoint E;
Savepoint created.
SQL> insert into class1 values(11,'sam');
1 row created.
SQL> savepoint F;
Savepoint created.
SQL> rollback to E;
Rollback complete.
ID NAME
---------- ----------
5 raj
6 ram
7 vibhav
8 sai
9 siva
10 tom

6 rows selected.
SQL> commit;
Commit complete.
ID NAME
---------- ----------
5 raj
6 ram
7 vibhav
8 sai
9 siva
10 tom
6 rows selected.
SQL> rollback to F;
rollback to F
*
ERROR at line 1:
ORA-01086: savepoint 'F' never established
SQL>
Output:

Experiment 4
AIM:To write a program on Inbuilt functions in SQL on sample exercise.
Program:
SQL> create table stu1(id number(10),name varchar(10),department
varchar(10),mark1 number(10),mark2 number(10),mark3 number(10));
Table created.
SQL> insert into stu1
values(&id,'&name','&department',&mark1,&mark2,&mark3);
Enter value for name: aaa
Enter value for department: cse
Enter value for mark1: 90
old 1: insert into stu1
values(&id,'&name','&department',&mark1,&mark2,&mark3)
new 1: insert into stu1 values(100,'aaa','cse',90,89,95)
1 row created.
SQL> /
Enter value for name: bbb

new 1: insert into stu1 values(101,'bbb','cse',88,89,90)
1 row created.
SQL> /
Enter value for name: ccc
new 1: insert into stu1 values(102,'ccc','cse',90,88,87)
1 row created.
SQL> /

Enter value for name: ddd
new 1: insert into stu1 values(103,'ddd','cse',75,80,85)
1 row created.
SQL> select * from stu1;
ID NAME DEPARTMENT MARK1 MARK2 MARK3
---------- ---------- ---------- ---------- ---------- ----------
100 aaacse 90 89 95
101 bbbcse 88 89 90
102 ccccse 90 88 87
103 dddcse 75 80 85
SQL> select count(*) from stu1;
COUNT(*)
----------
4

SQL> insert into stu1
values(&id,'&name','&department',&mark1,&mark2,&mark3);
Enter value for name: eee
Enter value for department: ece
new 1: insert into stu1 values(104,'eee','ece',88,87,88)
1 row created.
SQL> select count(*) from stu1;
COUNT(*)
----------
5
SQL> select count(*) from stu1 where department='ece';
COUNT(*)
----------

1
SQL> select count(*) from stu1 where department='cse';
COUNT(*)
----------
4
SQL> select count(id) from stu1;
COUNT(ID)
----------
5
SQL> select count(id) from stu1 where department='cse';
COUNT(ID)
----------
4
SQL> select count(id) from stu1 where department='ece';
COUNT(ID)
----------

1
SQL> select min(mark1) from stu1;
MIN(MARK1)
----------
75
MIN(MARK2)
----------
80
MIN(MARK3)
----------
85
SQL> select min(mark1+mark2) from stu1;
MIN(MARK1+MARK2)
----------------

155
MIN(MARK2+MARK3)
----------------
165
MIN(MARK1+MARK3)
----------------
160
SQL> select max(mark1) from stu1;
MAX(MARK1)
----------
90
MAX(MARK2)
----------

89
MAX(MARK3)
----------
95
SQL> select max(mark1+mark2) from stu1;
MAX(MARK1+MARK2)
----------------
179
MAX(MARK2+MARK3)
----------------
184
MAX(MARK1+MARK3)
----------------

185
SQL> select max(mark2) from stu1 where department='cse';
MAX(MARK2)
----------
89
SQL> select min(mark1) from stu1 where department='cse';
MIN(MARK1)
----------
75
SQL> select avg(mark1) from stu1;
AVG(MARK1)
----------
86.2
AVG(MARK2)
----------

86.6
AVG(MARK3)
----------
89
SQL> select avg(mark3) from stu1 where department='cse';
AVG(MARK3)
----------
89.25
SQL> select sum(mark1) from stu1;
SUM(MARK1)
----------
431
SUM(MARK2)
----------

433
SUM(MARK3)
----------
445
SQL> select sum(mark1+mark2) from stu1;
SUM(MARK1+MARK2)
----------------
864
SUM(MARK1+MARK3)
----------------
876
SUM(MARK2+MARK3)
----------------

878
SQL> select sum(mark3) from stu1 where department='cse';
SUM(MARK3)
----------
357
SQL>
Output:

Experiment 5
AIM: To Construct a ER Model for the application to be constructed to a
Database
Steps for drawing ER Diagram:
1. First, identify the entities in your database. In this case, we have three
entities.
2. The second step involves identifying the relationships between the
selected entities.
3. The third step involves identifying cardinalities.
4. The fourth step is identifying entity attributes. Make sure that every
attribute is mapped to only one entity; assign modifiers for those that
belong to more than one.
5. Once you have identified the entities, relationships, cardinalities, and
attributes, you can now create your ER diagram. Here’s what our sample
project will look like when designed using the crow’s foot (IE) notation.
6. Entity: Entities are represented by rectangle. All table of database are
treating as entity.
7. Attributes: Attributes are represented by ellipses. Attributes are
properties of entities.

(Link to draw ER diagram: https://app.diagrams.net)
Output:
Result: Hence we have successfully drawn the ER diagram.

Experiment: 6
AIM: To write a program for Nested Queries on sample exercise
Program :
SQL> select * from emp;
no rows selected
SQL> desc emp;
Name Null? Type
----------------------------------------- -------- ----------------------------
ID NUMBER(10)
NAME VARCHAR2(10)
AGE NUMBER(10)
ADDRESS VARCHAR2(10)
SALARY NUMBER(10)
SQL> insert into emp values(&id,'&name',&age,'&address',&salary);
Enter value for name: ram
Enter value for age: 33
Enter value for address: chennai
Enter value for salary: 50000
old 1: insert into emp values(&id,'&name',&age,'&address',&salary)
new 1: insert into emp values(10,'ram',33,'chennai',50000)
1 row created.
SQL> /
Enter value for name: raj
new 1: insert into emp values(20,'raj',20,'chennai',20000)
1 row created.

SQL> /
Enter value for name: sai
Enter value for address: mumbai
new 1: insert into emp values(30,'sai',26,'mumbai',15000)
1 row created.
SQL> /
Enter value for name: sam
Enter value for address: hyderabad
new 1: insert into emp values(40,'sam',27,'hyderabad',20000)
1 row created.
SQL> /
Enter value for name: tom
Enter value for address: pune
new 1: insert into emp values(50,'tom',29,'pune',65000)
1 row created.
SQL> /
Enter value for name: jerry

Enter value for address: kochin
new 1: insert into emp values(60,'jerry',30,'kochin',85000)
1 row created.
ID NAME AGE ADDRESS SALARY
---------- ---------- ---------- ---------- ----------
10 ram 33 chennai 50000
20 raj 20 chennai 20000
30 sai 26 mumbai 15000
40 sam 27 hyderabad 20000
50 tom 29 pune 65000
60 jerry 30 kochin 85000
6 rows selected.
---------- ---------- ---------- ---------- ----------
50 tom 29 pune 16250
60 jerry 30 hyderabad 25250
70 jack 33 bangalore 60000
80 jill 38 bangalore 65000
90 donald 23 chennai 10000
SQL> select max(salary) from emp;
MAX(SALARY)

-----------
65000
SQL> select id,name from emp where salary=65000;
ID NAME
---------- ----------
80 jill
SQL> select id,name from emp where salary=(select max(salary) from emp);
ID NAME
---------- ----------
80 jill
SQL> select * from emp where salary>(select avg(salary) from emp);
---------- ---------- ---------- ---------- ----------
SQL> select * from emp where salary>(select avg(salary) from emp)order by
age desc;
---------- ---------- ---------- ---------- ----------
SQL> select * from emp where salary<(select avg(salary) from emp)order by
age desc;
---------- ---------- ---------- ---------- ----------
50 tom 29 pune 16250

SQL> select * from emp where salary<=(select avg(salary) from emp);
---------- ---------- ---------- ---------- ----------
50 tom 29 pune 16250
7 rows selected.
SQL> select * from emp where salary>=(select avg(salary) from emp);
---------- ---------- ---------- ---------- ----------
SQL> insert into emp values(100,'gow',30,'chennai',27100);
1 row created.
SQL> select * from emp where salary<=(select avg(salary) from emp);
---------- ---------- ---------- ---------- ----------
50 tom 29 pune 16250
100 gow 30 chennai 27100

8 rows selected.
SQL> select * from emp where salary>=(select avg(salary) from emp);
---------- ---------- ---------- ---------- ----------
SQL> select * from emp where id in(select id from emp where salary>20000);
---------- ---------- ---------- ---------- ----------
50 tom 29 pune 65000
SQL> select * from emp where id in(select id from emp where salary>=20000);
---------- ---------- ---------- ---------- ----------
50 tom 29 pune 65000
SQL> update emp set salary=salary*0.25 where age in(select age from emp
where age>=29);
3 rows updated.
---------- ---------- ---------- ---------- ----------

50 tom 29 pune 16250
6 rows selected.
SQL> update emp set salary=salary*10 where age not in(select age from emp
where age>=30);
5 rows updated.
---------- ---------- ---------- ---------- ----------
30 sai 26 mumbai 150000
50 tom 29 pune 162500
SQL> select * from emp where salary<any(select max(salary) from emp);
---------- ---------- ---------- ---------- ----------
30 sai 26 mumbai 150000
50 tom 29 pune 162500

SQL> select * from emp where salary=all(select salary from emp where
salary<=20000);
---------- ---------- ---------- ---------- ----------
SQL> select * from emp where salary>all(select salary from emp where
salary<=20000);
---------- ---------- ---------- ---------- ----------
30 sai 26 mumbai 150000
50 tom 29 pune 162500
SQL> delete from emp where age in(select age from emp where age>=30);
2 rows deleted.
---------- ---------- ---------- ---------- ----------
50 tom 29 pune 16250

VIEWS
SQL> connect
Enter password:
Connected.
SQL> create taable s(sid number(10),sname varchar(10),address varchar(10));
create taable s(sid number(10),sname varchar(10),address varchar(10))
*
ERROR at line 1:
ORA-00901: invalid CREATE command
SQL> create table s(sid number(10),sname varchar(10),address varchar(10));
Table created.
SQL> insert into s values(1,'ram','chennai');
1 row created.
SQL> insert into s values(2,'raj','chennai');
1 row created.

SQL> insert into s values(3,'siva','pune');
1 row created.
SQL> insert into s values(4,'selvam','bangalore');
1 row created.
SQL> select * from s;
SID SNAME ADDRESS
---------- ---------- ----------
1 ram chennai
2 raj chennai
3 siva pune
4 selvam bangalore
SQL> create table s1(sid number(10),sname varchar(10),marks number(10),age
varchar(10));
Table created.
SQL> insert into s1 values(1,'ram',90,20);

1 row created.
SQL> insert into s1 values(2,'raj',90,21);
1 row created.
SQL> insert into s1 values(3,'siva',80,19);
1 row created.
SQL> insert into s1 values(4,'selvam',88,19);
1 row created.
SQL> select * from s1;
SID SNAME MARKS AGE
---------- ---------- ---------- ----------
1 ram 90 20
2 raj 90 21
3 siva 80 19
4 selvam 88 19
SQL> create view sv as select sname,address from s where sid<4;

View created.
SQL> select * from sv;
SNAME ADDRESS
---------- ----------
ram chennai
raj chennai
siva pune
SQL> create view sv1 as select s.sname,s.address,s1.marks from s,s1 where
s.sid=s1.sid;
View created.
SQL> select * from sv1;
SNAME ADDRESS MARKS
---------- ---------- ----------
ram chennai 90
raj chennai 90
siva pune 80
selvam bangalore 88

SQL> drop view sv1;
View dropped.
SQL> select * from sv1;
select * from sv1
*
ERROR at line 1:
Output:
Result: Hence the above query has been implemented successfully.

Experiment 7
AIM: To write a program for Join Queries on sample exercise.
Program:
SQL> create table fac(fid number(10),fname varchar(10),address
varchar(10),age number(10));
Table created.
SQL> create table fcou(cid number(10),fid number(10));
Table created.
SQL> insert into fac values(&fid,'&fname','&address',&age);
Enter value for fid: 1
Enter value for fname: aaa
old 1: insert into fac values(&fid,'&fname','&address',&age)
new 1: insert into fac values(1,'aaa','chennai',30)
1 row created.
SQL> /
Enter value for fname: bbb

new 1: insert into fac values(2,'bbb','chennai',33)
1 row created.
SQL> /
Enter value for fname: ccc
Enter value for address: bangalore
new 1: insert into fac values(3,'ccc','bangalore',34)
1 row created.
SQL> /
Enter value for fname: ddd
Enter value for address: kochin
new 1: insert into fac values(4,'ddd','kochin',30)
1 row created.

SQL> /
Enter value for fname: eee
new 1: insert into fac values(5,'eee','hyderabad',30)
1 row created.
SQL> /
Enter value for fname: fff
new 1: insert into fac values(6,'fff','hyderabad',29)
1 row created.
SQL> /
Enter value for fname: ggg

new 1: insert into fac values(7,'ggg','chennai',33)
1 row created.
SQL> select * from fac;
FID FNAME ADDRESS AGE
---------- ---------- ---------- ----------
1 aaa chennai 30
2 bbb chennai 33
3 ccc bangalore 34
4 ddd kochin 30
5 eee hyderabad 30
6 fff hyderabad 29
7 ggg chennai 33
7 rows selected.
SQL> insert into fcou values(10,1);
1 row created.

1 row created.
1 row created.
1 row created.
1 row created.
1 row created.
SQL> select * from fcou;
CID FID
---------- ----------

10 1
10 2
20 3
30 4
20 5
20 6
6 rows selected.
SQL> select * from fac;
FID FNAME ADDRESS AGE
---------- ---------- ---------- ----------
1 aaa chennai 30
2 bbb chennai 33
3 ccc bangalore 34
4 ddd kochin 30
5 eee hyderabad 30
6 fff hyderabad 29
7 ggg chennai 33
7 rows selected.
SQL> select fcou.cid,fac.fname,fac.age from fac inner join fcou on
fac.fid=fcou.fid;
CID FNAME AGE

---------- ---------- ----------
10 aaa 30
10 bbb 33
20 ccc 34
30 ddd 30
20 eee 30
20 fff 29
6 rows selected.
SQL> select fac.fname,fcou.cid from fac left join fcou on fcou.fid=fac.fid;
FNAME CID
---------- ----------
aaa 10
bbb 10
ccc 20
ddd 30
eee 20
fff 20
ggg
7 rows selected.

SQL> select fac.fid,fcou.cid from fac left join fcou on fcou.fid=fac.fid;
FID CID
---------- ----------
1 10
2 10
3 20
4 30
5 20
6 20
7
7 rows selected.
SQL> select fac.fid,fcou.cid from fac right join fcou on fcou.fid=fac.fid;
FID CID
---------- ----------
1 10
2 10
3 20
4 30
5 20
6 20

6 rows selected.
1 row created.
1 row created.
SQL> select fac.fid,fcou.cid from fac right join fcou on fcou.fid=fac.fid;
FID CID
---------- ----------
1 10
2 10
3 20
4 30
5 20
6 20
7 40
40

8 rows selected.
SQL> select fac.fid,fcou.cid from fac full join fcou on fcou.fid=fac.fid;
FID CID
---------- ----------
1 10
2 10
3 20
4 30
5 20
6 20
7 40
40
8 rows selected.
SQL> select fcou.fid,fcou.cid from fac full join fcou on fcou.fid=fac.fid;
FID CID
---------- ----------
1 10
2 10
3 20

4 30
5 20
6 20
7 40
8 40
8 rows selected.
SQL> select fac.fid,fcou.fid from fac full join fcou on fcou.fid=fac.fid;
FID FID
---------- ----------
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8
8 rows selected.
Output:
Result: Hence the above query has been implemented successfully.

Experiment: 8
AIM: To implement the program for Set Operators & Views.
Program:
SQL> create table z(id number(10),name varchar(10));
Table created.
SQL> insert into z values(1,'a');
1 row created.
SQL> insert into z values(2,'b');
1 row created.
SQL> create table z1(id number(10),name varchar(10));
Table created.
SQL> insert into z1 values(2,'b');
1 row created.

SQL> insert into z1 values(3,'c');
1 row created.
SQL> insert into z1 values(4,'d');
1 row created.
SQL> select * from z;
ID NAME
---------- ----------
1 a
2 b
SQL> select * from z1;
ID NAME
---------- ----------
2 b
3 c
4 d

SQL> select * from z union select * from z1;
ID NAME
---------- ----------
1 a
2 b
3 c
4 d
SQL> select * from z union all select * from z1;
ID NAME
---------- ----------
1 a
2 b
2 b
3 c
4 d
SQL> select * from z intersect select * from z1;

ID NAME
---------- ----------
2 b
SQL> select * from z minus select * from z1;
ID NAME
---------- ----------
1 a
SQL> select * from z1 minus select * from z;
ID NAME
---------- ----------
3 c
4 d
OUTPUT:
Result: The above query has been implemented successfully

Experiment: 9
AIM: To write a program on PL/SQL Conditional and Iterative Statements
Program:
Set serveroutput on
SQL> declare
2 a integer:=10;
3 begin
4 dbms_output.put_line(a);
5 end;
6 /
10
PL/SQL procedure successfully completed.
SQL> declare
2 a varchar(10):='hello';
3 begin
4 dbms_output.put_line(a);
5 end;
6 /
hello
SQL> declare
2 a integer:=10;

3 begin
4 dbms_output.put_line('value of a:'||a);
5 end;
6 /
value of a:10
SQL> declare
2 pi constant number:=3.14;
3 begin
4 dbms_output.put_line(pi);
5 end;
6 /
3.14
SQL> declare
2 a integer:=10;
3 b integer:=20;
4 c integer;
5 begin
6 c:=a+b;
7 dbms_output.put_line(c);
8 end;

9 /
30
IF-ELSE
SQL> declare
2 a integer:=30;
3 begin
4 if(a<20) then
5 dbms_output.put_line('a is less than 20');
6 else
7 dbms_output.put_line('a is not less than 20');
8 end if;
9 end;
10 /
a is not less than 20
CASE STATEMENT
SQL> DECLARE
2 grade char(1) := 'A';
3 BEGIN
4 CASE grade
5 when 'A' then dbms_output.put_line('Excellent');
6 when 'B' then dbms_output.put_line('Very good');
7 when 'C' then dbms_output.put_line('Good');

8 when 'D' then dbms_output.put_line('Average');
9 when 'F' then dbms_output.put_line('Passed with Grace');
10 else dbms_output.put_line('Failed');
11 END CASE;
12 END;
13 /
Excellent
LOOP
SQL> DECLARE
2 i NUMBER := 1;
3 BEGIN
4 LOOP
5 EXIT WHEN i>10;
6 DBMS_OUTPUT.PUT_LINE(i);
7 i := i+1;
8 END LOOP;
9 END;
10 /
1
2
3
4
5

6
7
8
9
10
WHILE LOOP
SQL> DECLARE
2 i INTEGER := 1;
3 BEGIN
4 WHILE i <= 10 LOOP
5 DBMS_OUTPUT.PUT_LINE(i);
6 i := i+1;
7 END LOOP;
8 END;
9 /
1
2
3
4
5
6
7
8

9
10
FOR LOOP
SQL> DECLARE
2 VAR1 NUMBER;
3 BEGIN
4 VAR1:=10;
5 FOR VAR2 IN 1..10
6 LOOP
7 DBMS_OUTPUT.PUT_LINE (VAR1*VAR2);
8 END LOOP;
9 END;
10 /
10
20
30
40
50
60
70
80
90
100

OUTPUT:
Result: The above query has been implemented successfully.

Experiment: 10
AIM: To write a program on PL/SQL Procedures on sample exercises.
Program:
SQL> create table proc(id number(10),name varchar(10));
Table created.
SQL> create or replace procedure "INSERTUSER"
(id IN NUMBER,
name IN VARCHAR2)
is
begin
insert into proc values(id,name);
end;
/
Procedure created.
SQL> BEGIN
insertuser(101,'Rahul');
dbms_output.put_line('record inserted successfully');
END;
/
record inserted successfully

SQL> select * from proc;
ID NAME
---------- ----------
101 Rahul
OUTPUT:

Experiment: 11
AIM: To write a program on PL/SQL Functions
Program:
SQL> create or replace function adder(n1 in number, n2 in number)
2 return number
3 is
4 n3 number(8);
5 begin
6 n3 :=n1+n2;
7 return n3;
8 end;
9 /
Function created.
SQL> DECLARE
2 n3 number(2);
3 BEGIN
4 n3 := adder(11,22);
5 dbms_output.put_line('Addition is: ' || n3);
6 END;
7 /
Addition is: 33

SQL> DECLARE
2 a number;
3 b number;
4 c number;
5 FUNCTION findMax(x IN number, y IN number)
6 RETURN number
7 IS
8 z number;
9 BEGIN
10 IF x > y THEN
11 z:= x;
12 ELSE
13 Z:= y;
14 END IF;
15
16 RETURN z;
17 END;
18 BEGIN
19 a:= 23;
20 b:= 45;
21
22 c := findMax(a, b);
23 dbms_output.put_line(' Maximum of (23,45): ' || c);

24 END;
25 /
Maximum of (23,45): 45
---------- ---------- ---------- ---------- ----------
30 sai 26 mumbai 150000
50 tom 29 pune 40630
SQL> create or replace function totalemp
2 return number is
3 total number(10):=0;
4 begin
5 select count(*) into total from emp;
6 return total;
7 end;
8 /
Function created.

SQL> declare
2 c number(20);
3 begin
4 c:=totalemp();
5 dbms_output.put_line('Total no of emp:'||c);
6 end;
7 /
Total no of emp:3
SQL> DECLARE
2 num number;
3 factorial number;
4
5 FUNCTION fact(x number)
6 RETURN number
7 IS
8 f number;
9 BEGIN
10 IF x=0 THEN
11 f := 1;
12 ELSE
13 f := x * fact(x-1);
14 END IF;

15 RETURN f;
16 END;
17
18 BEGIN
19 num:= 6;
20 factorial := fact(num);
21 dbms_output.put_line(' Factorial '|| num || ' is ' || factorial);
22 END;
23 /
Factorial 6 is 720
OUTPUT:

Experiment: 12
AIM: To write program on PL/SQL Cursors
Program:
SQL> DECLARE
total_rows number(2);
BEGIN
UPDATE emp
SET salary = salary + 5000;
IF sql%notfound THEN
dbms_output.put_line('no customers updated');
ELSIF sql%found THEN
total_rows := sql%rowcount;
dbms_output.put_line( total_rows || ' customers updated ');
END IF;
END;
/
3 customers updated
---------- ---------- ---------- ---------- ----------

30 sai 26 mumbai 155000
50 tom 29 pune 45630
EXPLICIT CURSOR
SQL> DECLARE
c_id emp.id%type;
c_name emp.name%type;
c_addr emp.address%type;
CURSOR c_emp is
SELECT id, name, address FROM emp;
begin
open c_emp;
loop
FETCH c_emp into c_id, c_name, c_addr;
EXIT WHEN c_emp%notfound;
dbms_output.put_line(c_id || ' ' || c_name || ' ' || c_addr);
end loop;
close c_emp;
end;
/
30 sai mumbai
40 sam hyderabad
50 tom pune

Experiment: 13
AIM: To write program on PL/SQL Exception Handling
Program:
SQL> DECLARE
c_id emp.id%type:=10;
begin
SELECT name, address into c_name,c_addr FROM emp where id=c_id;
dbms_output.put_line('Name:'||c_name);
dbms_output.put_line('Address:'||c_addr);
exception
when no_data_found then dbms_output.put_line('no such customer');
when others then dbms_output.put_line('error');
end;
/
no such customer
SQL> declare
c_id emp.id%type:=30;

begin
SELECT name, address into c_name,c_addr FROM emp where id=c_id;
dbms_output.put_line('Name:'||c_name);
dbms_output.put_line('Address:'||c_addr);
exception
end;
/
Name:sai
Address:mumbai
USER DEFINED
SQL> declare
c_id emp.id%type:=&id;
ex_invalid_id exception;
begin
if c_id<=0 then
raise ex_invalid_id;
else
select name,address into c_name,c_addr from emp where id=c_id;

dbms_output.put_line('name:'||c_name);
dbms_output.put_line('address:'||c_addr);
end if;
exception
when ex_invalid_id then dbms_output.put_line('id should be greater than zero');
end;
/
Enter value for id: -6
old 2: c_id emp.id%type:=&id;
new 2: c_id emp.id%type:=-6;
id should be greater than zero
SQL> /
old 2: c_id emp.id%type:=&id;
new 2: c_id emp.id%type:=30;
name:sai
address:mumbai

OUTPUT:
Result: The above query has been implemented successfully.

Experiment 14
AIM: To write a program on PL/SQL Trigger
Program:
SQL> CREATE OR REPLACE TRIGGER display_salary_changes
BEFORE DELETE OR INSERT OR UPDATE ON emp
FOR EACH ROW
WHEN (NEW.ID > 0)
DECLARE
sal_diff number;
BEGIN
sal_diff := :NEW.salary - :OLD.salary;
dbms_output.put_line('Old salary: ' || :OLD.salary);
dbms_output.put_line('New salary: ' || :NEW.salary);
dbms_output.put_line('Salary difference: ' || sal_diff);
END;
/
Trigger created.
SQL> DECLARE
total_rows number(2);
BEGIN
UPDATE emp
SET salary = salary + 5000;
IF sql%notfound THEN

dbms_output.put_line('no customers updated');
ELSIF sql%found THEN
total_rows := sql%rowcount;
dbms_output.put_line( total_rows || ' customers updated ');
END IF;
END;
/
Old salary: 155000
New salary: 160000
Salary difference: 5000
Old salary: 205000
New salary: 210000
Old salary: 45630
New salary: 50630
3 customers updated
Output:
Result: Hence the above program is implemented successfully.

Experiment 15
AIM: To make a project with front-end and back-end using oracle DBMS
Procedure:
1. Consider the database of at least 4 schemas/tables.
2. Plan for attributes that cover all general data types.
3. Plan for constraints of different types (primary keys, foreign keys, unique
keys, check, not null etc) with appropriate constraint names.
4. Create appropriate data for the above schemas.
5. Implement the database in Oracle 10g Express Edition creating a user.
6. Plan queries and find the answers (at least 10).
7. Formulate everything in a report which includes:
o Brief description of the system that you are going to implement in
the database
o Mention schemas with attributes
o Expected functional dependencies on your schemas.
o Proof of good database design so that the schemas are in good form
(BCNF or 3NF)
o Schema diagram of the database
o E-R diagram of the database
o Snapshots of SQL DDL of all the schemas/tables
o Snapshots of the instances (data of the populated tables)
o Query statements in English language, SQL statements and
snapshots of the outputs. SQL statements should contain the
following :
o natural join, cross product, outer join, join with using, on
o Nested sub-queries with clauses (some, all, any, exists, unique etc.)
o order by, group by, having clauses
o Use of with clause
o Use of Functions (string/text, numeric) , set operations (union,
intersect, difference/minus)

o Exception handling and cursor operations
o Update, delete operations
 Conclusion of the work
Output:
Result: Hence the project was implemented successfully.

"Comprehensive Guide to Database Management Systems: Optimizing Data Architecture, Administration, and Performance for Enhanced Organizational Efficiency and Decision-Making"

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