plsqlprograms

http://placementkit.blogspot.com/

PL / SQL PROGRAMS P1. WRITE A NESTED PROGRAM TO ADD AND TO MULTIPLY TWO NUMBERS. DECLARE N1 number; N2 number; Sum number ; BEGIN Sum := N1+N2 ; > DECLARE Prod number; BEGIN Prod := N1 * N2 ; Dbms_output.put_line( Product Value = prod ); END inner_block ; Dbms_output.put_line( Sum Value = sum); END; P2. WRITE A PROGRAM TO CALCULATE THE SIMPLE INTEREST AND COMPUND INTEREST , IF P, N, R ARE GIVEN. declare p number(9,2) ; n number(9,2) ; r number(9,2) ; si number(9,2) := 0; ci number(9,2) := 0; begin p := &principal_amount; n := &no_of_years; r := &rate_of_interest; si := p*n*r/100; ci := p*(1+r/100)**n; dbms_output.put_line('simple interset =' si); dbms_output.put_line('compound interset =' ci); end; SQL> / Enter value for principal_amount: 10000 old 8: p:=&principal_amount; new 8: p:=10000; Enter value for no_of_years: 5 old 9: n:=&no_of_years; new 9: n:=5; Enter value for rate_of_interest: 10.5 old 10: r:=&rate_of_interest; new 10: r:=10.5; simple interset =5250 compound interset =16474.47 PL/SQL procedure successfully completed. PROGRAM BASED ON IF LOOP


P3 . Write a program to check greatest of two numbers : declare a number(3) :=20; b number(3) :=10; begin if a>b then dbms_output.put_line('a is the greatest : ' a); else dbms_output.put_line('B is the greatest : ' b); end if; end; SQL> / a is the greatest : 20 PL/SQL procedure successfully completed. P4. Given 2 sides of a rectangle .Write a program to find out its area is gr eater than its perimeter or not . declare l number; b number; ar number; pr number; begin l := &l; b := &b; ar := l*b; pr := 2*(l+b); if ar > pr then dbms_output.put_line('the area iS > its perimeter' 'area = ' ar 'perimet er = ' pr); else dbms_output.put_line('the area iS < its perimeter' ' area = ' ar ' perim eter = ' pr); end if; end; Enter value for l: 10 old 7: l:=&l; new 7: l:=10; Enter value for b: 6 old 8: b:=&b; new 8: b:=6; the area is > its perimeter area = 60 perimeter = 32 PL/SQL procedure successfully completed. P5. WRITE A PROGRAM TO INPUT A SINGLE DIGIT NO: CONVERT IT INTO WORDS. Declare a number; t varchar2(10); begin a :=&a; if a=1 then


t := 'one'; elsif a=2 then t := 'two'; elsif a= 3 then t := 'three'; elsif a= then t := 'four'; elsif a=5 then t := 'five'; elsif a=6 then t := 'six'; elsif a=7 then t := 'seven'; elsif a=8 then t := 'eight'; elsif a=9 then t := 'nine'; else t := 'zero'; end if; dbms_output.put_line(a ' = ' t); end; Enter value for a: 2 old 5: a:=&a; new 5: a:=2; 2 = two PL/SQL procedure successfully completed. P6 . Write a program to check the given number is +ve or ve : SQL> declare n number(2):=12; begin if n>0 then dbms_output.put_line('the given number is positive' n); else dbms_output.put_line('the given number is negative' n); end if; end; / SQL> save posneg.sql Created file posneg.sql SQL >The given number is positive 12 PL/SQL procedure successfully completed. PROGRAM BASED ON NESTED IF LOOP P7. WRITE A PROGRAM TO INPUT 2 NUMBERS IF THE 1st No >2nd No THEN SWAP IT, ELSE IF 1st No < 2nd No RAISE IT TO ITS POWER , ELSE DOUBLES IT. declare a number(5); b number(5);


t number(5); begin a := &a; b := &b; dbms_output.put_line( 'initial value of a = ' a ' b = ' b ) if a>b then t:= a; a:= b; b:=t; elsIF A / Enter value for a: 4 old 6: a:=&a; new 6: a:=4; Enter value for b: 5 old 7: b:=&b; new 7: b:=5; initial value of a = 4 b = 5 final value of a = 256 b = 3125 PL/SQL procedure successfully completed. SQL> / Enter value for a: 5 old 6: a:=&a; new 6: a:=5; Enter value for b: 4 old 7: b:=&b; new 7: b:=4; initial value of a = 5 b = 4 final value of a = 4 b = 5 PL/SQL procedure successfully completed. SQL> Enter value for a: 5 old 6: a:=&a; new 6: a:=5; Enter value for b: 5 old 7: b:=&b; new 7: b:=5; initial value of a = 5 b = 5 final value of a = 10 b = 10 PL/SQL procedure successfully completed. P8. A bank accepts fixed deposits for one or more years and the policy it ad


opts on interest is as follows: If a deposit is < Rs 2000 and for 2 or more years , the interest rat e is 5% compounded annually. If a deposit is Rs.2000 or more but less than 6000 and for 2 or more years , the interest is 7 % compounded annually. If a deposit is Rs.6000 and for 1 or more years , the interest is 8 % compounded annually. On all deposits for 5 years or more , interest is 10% compounded ann ually. On all other deposits not covered by above conditions , the interest is 3% compounded annually. Given the amount deposited and the number of years , Write a program to calc ulate the amount on maturity. declare p number(9,2); r number(9,2); t number(9,2); ci number(9,2); begin p := &p; t := &t; if p=2 then r := 5; elsif p>=2000 and p=2 then r := 7; elsif p>6000 and t>=1 then r := 8; elsif t>=5 then r := 10; else r := 3; end if; ci := p*(1+r/100)**t - p; dbms_output.put_line('The Principal amount is =' p); dbms_output.put_line('The rate of interest is =' r); dbms_output.put_line('The time period is =' t); dbms_output.put_line('The compund interst is =' ci); end; Enter value for p: 10000 old 7: p:=&p; new 7: p:=10000; Enter value for t: 5 old 8: t:=&t; new 8: t:=5; The Principal amount is =10000 The rate of interest is =8 The time period is =5 The compund interst is =4693.28 PL/SQL procedure successfully completed. Enter value for p: 1500


old 7: p:=&p; new 7: p:=1500; Enter value for t: 3 old 8: t:=&t; new 8: t:=3; The Principal amount is =1500 The rate of interest is =5 The time period is =3 The compound interest is =236.44 PL/SQL procedure successfully completed. P9. WRITE A PROGRM TO INPUT 3 NUMBERS FIND THE 1st Greatest, 2nd Greatest, 3rd Greatest. declare a number(3); b number(3); c number(3); f number(3); s number(3); t number(3); begin a :=&a; b :=&b; c :=&c; if a>b and a>c then f:= a; if b>c then s:=b; t:=c; else s:=c; t:=b; end if; elsif b>a and b>c then f:= b; if a>c then s:=a; t:=c; else s:=c; t:=a; end if; else f:= c; if a>b then s:=a; t:=b; else s:=b; t:=a; end if; end if; dbms_output.put_line('first largest = ' f);


dbms_output.put_line('second largest = ' s); dbms_output.put_line('third largest = ' t); end; / Enter value for a: 5 old 9: a:=&a; new 9: a:=5; Enter value for b: 8 old 10: b:=&b; new 10: b:=8; Enter value for c: 9 old 11: c:=&c; new 11: c:=9; first largest = 9 second largest = 8 third largest = 5 PL/SQL procedure successfully completed. P10 . WRITE A PROGRAM TO INPUT 2 NUMBERS AND AN OPERATOR , AND DISPLAY THE RESULT. declare a number(3) ; b number(3) ; c number(3) ; op char(1) ; begin a := &a ; b := &b ; op := &op ; if op='+' then c:=a+b; elsif op='-' then c:=a-b; elsif op='*' then c:=a*b; else c:=a/b; end if; dbms_output.put_line('result=' c); end; Enter value for a: 5 old 7: a:=&a; new 7: a:=5; Enter value for b: 6 old 8: b:=&b; new 8: b:=6; Enter value for op: '*' old 9: op:=&op;


new 9: op:='*'; result=30 PL/SQL procedure successfully completed. P11. Write a program to calculate the commission of the sales man. If salesmade Comm >10000 500 10000 20000 1000 >20000 1500 declare sman varchar(10); sm number(9,2); com number(9,2); begin sman := &sman; sm := &sm; if sm > 10000 then com := 500; elsif sm > 20000 then com := 1000; else com := 1500; end if; dbms_output.put_line(' the sales man name is :' sman); dbms_output.put_line(' the sales made is :' sm); dbms_output.put_line(' the sales commission is :' com); end; Enter value for sman: 'Mathew' old 6: sman:=&sman; new 6: sman:='Mathew'; Enter value for sm: 15000 old 7: sm:=&sm; new 7: sm:=15000; The sales man name is :Mathew The sales made is :15000 The sales commission is :500 PL/SQL procedure successfully completed. PROGRAM BASED ON WHILE LOOP P12. TO GENERATE NUMBERS FROM 0 TO 25 IN STEP OF 5 declare i number :=10 ; begin dbms_output.put_line(' THE WHILE LOOP begins'); WHILE I


20 25 PL/SQL procedure successfully completed. P13. Write a program to find the sum of the digits of the number: DECLARE N number ; S NUMBER :=0; R NUMBER; begin n:=&N; WHILE N0 LOOP R := MOD(N,10); S := S + R; N := TRUNC(N/10); end loop; dbms_output.put_line('THE SUM OF THE DIGITS = ' S); end; SQL> Enter value for n: 375 old 7: n:=&N; new 7: n:=375; THE SUM OF THE DIGITS = 15 PL/SQL procedure successfully completed. PROGRAM BASED ON NESTED WHILE LOOP PROGRAM BASED ON FOR LOOP P14. WRITE A PROGRAM CODE TO PRINT THE MULTIPLICATION TABLE OF A GIVEN NO: declare t number(3) := 3; begin T := &T; FOR I IN 1..3 LOOP dbms_output.put_line(t ' X ' i ' = ' i*t ); end loop; end; p15. write aprogram to generate even numbers from 2 to 50, and find its sum. declare i number(5); n number(5); v number(5); s number(5):=0; begin n := &terminal_number; for i in 1 .. n/2 loop v := i*2; s := s+v; dbms_output.put_line(v); end loop; dbms_output.put_line('the sum of numbers from 2 to ' n ' = ' s); end; SQL> /


Enter value for terminal_number: 10 old 7: n:=&terminal_number; new 7: n:=10; 2 4 6 8 10 The sum of numbers from 2 to 10 = 30 PL/SQL procedure successfully completed. P16 . WRITE A PROGRAM TO GENERATE FIRST 25 TERMS OF THE FIBONACCIS SERIES. declare a number:= 0 ; b number:= 1; c number; begin dbms_output.put(a ' ' b ' '); for i in 3..10 loop c := a + b; dbms_output.put(c ' '); a := b; b := c; end loop; dbms_output.put_line(' '); end; 0 1 1 2 3 5 8 13 21 34 PL/SQL procedure successfully completed. P17. Write a program to find the factorial of a number : declare n number(2); i number(2); f number(5):=1; begin n :=&n; for i in 1..n loop f := f * i; end loop; dbms_output.put_line(' the factorial value = ' f); end; Enter value for n: 5 old 6: n:=&n; new 6: n:=5; the factorial value = 120 PROGRAM BASED ON NESTED FOR LOOP P18. Write a program to print the following design:


1 12 123 1234 12345 declare i number ; j number; n number; begin n :=&n; for i in 1..n loop for j in 1..i loop dbms_output.put(j); end loop; dbms_output.put_line(' '); end loop; end; Enter value for n: 5 old 6: n:=&n; new 6: n:=5; P19. WRITE A PROGRAM TO DISPLAY NUMBERS OF THE FORM 0 0 0 0 0 1 2 3 4 5 2 4 6 8 10 3 6 9 12 15 4 8 12 16 20 5 10 15 20 25 DECLARE I NUMBER; J NUMBER ; K NUMBER; BEGIN FOR I IN 0 .. 5 LOOP FOR J IN 1..5 LOOP K := I*J; DBMS_OUTPUT.PUT(K); END LOOP; DBMS_OUTPUT. PUT_LINE (' '); END LOOP; END; P20. WRITE A PL/SQL CODE TO ACCEPT THE TEXT AND REVERSE THE GIVEN TEXT. CHECK THE TEXT IS PALINDROME OR NOT DECLARE G VARchar2(20); r VARchar2(20); BEGIN G:='&g'; dbms_output.put_line('THE GIVEN TEXT :' G); for i in REVERSE 1.. length(G) loop R:= R substr(G,i,1);


end loop; dbms_output.put_line('THE REVERSED TEXT :' R); IF R=G THEN dbms_output.put_line('THE GIVEN TEXT IS PALINDROME '); ELSE dbms_output.put_line('THE GIVEN TEXT IS NOT PALINDROME '); END IF; end; SQL> / Enter value for g: MALAYALAM old 5: G:='&g'; new 5: G:='MALAYALAM'; THE GIVEN TEXT :MALAYALAM THE REVERSED TEXT :MALAYALAM THE GIVEN TEXT IS PALINDROME PL/SQL procedure successfully completed. / Enter value for g: HELLO old 5: G:='&g'; new 5: G:='HELLO'; THE GIVEN TEXT :HELLO THE REVERSED TEXT :OLLEH THE GIVEN TEXT IS NOT PALINDROME PL/SQL procedure successfully completed. P21. Write a program to print the following design: declare i number ; j number; n number; k number; m number; begin n := &n; for i in 1..n loop for j in 1..n-i loop dbms_output.put('~'); end loop; for k in 1..i loop dbms_output.put(k); end loop; for k in reverse 1..i-1 loop dbms_output.put(k); end loop; dbms_output.put_line(' '); end loop; end; Enter value for n: 5 old 8: n:=&n; new 8: n:=5; ~~~~1 ~~~121 ~~12321


~1234321 123454321 PL/SQL procedure successfully completed. P22. Write a program to print the following design : declare i number ; j number; n number; k number; m number; begin n := &n; for i in 1..n loop for j in 1..n-i loop dbms_output.put('~'); end loop; for k in 1..i loop dbms_output.put(k); end loop; for k in reverse 1..i-1 loop dbms_output.put(k); end loop; dbms_output.put_line(' '); end loop; for i in reverse 1..n-1 loop for j in 1..n-i loop dbms_output.put('~'); end loop; for k in 1..i loop dbms_output.put(k); end loop; for k in reverse 1. . i-1 loop dbms_output.put(k); end loop; dbms_output.put_line(' '); end loop; end; Enter value for n: 5 old 8: n:=&n; new 8: n:=5; ~~~~1 ~~~121 ~~12321 ~1234321 123454321 ~1234321 ~~12321 ~~~121 ~~~~1 PL/SQL procedure successfully completed. PROGRAM BASED ON FOR & IF P23. GENERATE ODD NOS: FROM 1 TO 10 AND FIND ITS SUM.


DECLARE I NUMBER(4); S NUMBER(4):=0; BEGIN FOR I IN 1..10 LOOP IF MOD(I,2) 0 THEN S := S+I; DBMS_OUTPUT.PUT_LINE(I); END IF; END LOOP; DBMS_OUTPUT.PUT_LINE(' THE SUM OF ODD NOS FROM 1 TO 10 = ' S); END; SQL>/ 1 3 5 7 9 THE SUM OF ODD NOS FROM 1 TO 10 = 25 PL/SQL procedure successfully completed. P24 . WRITE A PROGRAM TO ALL ODD NUMBERS FORM 10 TO 1 IN REVERSE ORDER. declare i number(5); n number(5); v number(5); s number(5) :=0; begin S := &STARTING_NUMBER; n := &terminal_number; for i in REVERSE S .. N loop IF MOD (I,2) 0 THEN s := s + I; dbms_output.put_line(I); end loop; dbms_output.put_line('the sum of numbers from 2 to ' n ' = ' s); end; Enter value for starting number : 1 Old 6: s=&starting_number:1 New 6: s:=1; Enter value for Terminal_number: 10 old 7: n := &terminal_number; new 7: n:=10; 9 7 5 3 1 The sum of numbers from 1 to 9 = 25 PL / SQL procedure successfully completed. P25. Write a program to check the given no: is prime or not:


declare n number; i number; pr number(2):=1; begin n:= &n; for i in 2..n/2 loop if mod(n,i) = 0 then pr := 0; end if; end loop; if pr = 1 then dbms_output.put_line(' the given no: is prime: ' n); else dbms_output.put_line(' the given no: is not prime: ' n); end if; end; Enter value for n: 7 old 6: n:=&n; new 6: n:=7; the given no: is prime: 7 PL/SQL procedure successfully completed. SQL> / Enter value for n: 25 old 6: n:=&n; new 6: n:=25; the given no: is not prime: 25 PL/SQL procedure successfully completed. P26. WRITE A PROGRAM TO PRINT ASCII TABLE : DECLARE I NUMBER; BEGIN FOR I IN 33..256 LOOP DBMS_OUTPUT.PUT (( TO_CHAR (I,'000')) ':' CHR(I) ' ' ); IF MOD (I, 8) = 0 THEN DBMS_OUTPUT.PUT_LINE(' '); END IF; END LOOP; END; 033:! 034:" 035:# 036:$ 037:% 038:& 039:' 040:( 041:) 042:* 043:+ 044:, 045:- 046:. 047:/ 048:0 049:1 050:2 051:3 052:4 053:5 054:6 055:7 056:8 249: 250: 251: 252: 253: 254: 255: 256: :: :: :: :: :: :: :: :: :: :: :: :: PL/SQL procedure successfully completed. P27. Write a program to generate prime nos from 1 to 10:


declare n number; i number; pr number; begin for n in 1..10 loop pr:=1; for i in 2 .. n/2 loop if mod(n,i) = 0 then pr := 0; end if; end loop; if pr = 1 then dbms_output.put_line(n); end if; end loop; end; 1 2 3 5 7 P28 . Write a program to check the square root of a number is prime or not. declare n number; i number; pr number; begin pr := 1; n := &n; n := sqrt(n); for i in 2 .. n/2 loop if mod(n,i) = 0 then pr := 0; end if; end loop; if pr = 1 then dbms_output.put_line('the square root of the given number is prime' n*n); else dbms_output.put_line('the square root of the given number is not prime' n*n ); end if; end; SQL> Enter value for n: 81 old 7: n:=&n; new 7: n:=81; The square root of the given number is not prime81 PL/SQL procedure successfully completed. PROGRAM BASED ON LOOP... END LOOP P29. Write a program to reverse the digits of the number:


DECLARE N number ; S NUMBER : = 0; R NUMBER; K number; begin N := &N; K := N; LOOP EXIT WHEN N = 0 ; S := S * 10; R := MOD(N,10); S := S + R; N := TRUNC(N/10); end loop; dbms_output.put_line( ' THE REVERSED DIGITS ' ' OF ' K ' = ' S); end; Enter value for n: 4567 old 7: N:=&N; new 7: N:=4567; THE REVERSED DIGITS OF 4567 = 7654 PL/SQL procedure successfully completed. PROGRAM BASED ON RECORDS P30. WRITE PL/SQL SCRIPT TO CREATE A RECORD TYPE THAT CAN HOLD SALES IN 4 DIFFERENT QUARTERS. Structure of the table : SALES ( SNO, NAME, QUAD1, QUAD2, QUAD3, QUAD4 ) SALES2004 ( Q1, Q2, Q3, Q4 ) DECLARE TYPE SALEREC IS RECORD ( Q1 NUMBER, Q2 NUMBER, Q3 NUMBER, Q4 NUMBER, ) ; BEGIN SREC SALEREC ; SELECT SUM(QUAD1) , SUM(QUAD2) , SUM(QUAD3), SUM(QUAD4) INTO SREC FROM SALES ; INSERT INTO SALES2004 VALUES (SREC.Q1 ,SREC.Q2, SREC.Q3, SREC.Q4); END; PROGRAM BASED ON DATABASE INTERACTION P31 . Write PL/SQL SCRIPT to determine the salary of employee in the EMP tab le. If salary >7500 give an increment of 15% , otherwise display the message NO INCREMENT . Get the empno from the user. DECLARE ENO EMP.EMPNO%TYPE; SALARY EMP.SAL%TYPE; BEGIN ENO := &ENO;


SELECT SAL INTO SALARY FROM EMP WHERE EMPNO=ENO; IF SALARY >7500 THEN UPDATE EMP SET SAL =SAL+SAL* 15/100 WHERE EMPNO=ENO; ELSE DBMS_OUTPUT.PUT_LINE(NO INCREMENT) END IF; END; P32 . GIVEN A TABLE STUDENT ( SID INTEGER PRIMARY KEY, NAME CHAR(30), AGE INTEGER , GPA FLOAT ). WRITE A PL/SQL SCRIPT TO GO THROUGH SID 142-857 AND SET ALL GPA UNDER 4.0 TO 4.0 . DECLARE TSID STUDENT.SID%TYPE; TGPA STUDENT.GPA%TYPE; BEGIN TSID:=142; LOOP EXIT WHEN TSID > 857 ; SELECT GPA INTO TGPA FROM STUDENT WHERE SID=TSID; IF TGPA< 4.0 THEN UPDATE STUDENT SET GPA=4.0 WHERE SID=TSID; END IF; TSID:=TSID+1; END LOOP; END; P33 . DISPLAY AVERAGE SALARY , TOTOL NO: OF EMP, AND TOTAL SALARY FOR FIRST 5 DEPARTMENTS IN EMP TABLE. DEPT NO ARE 10,20,30,40,5090. DECLARE DNO NUMBER:=10; DPNAME DEPT. DNAME%TYPE; ASAL EMP. SAL%TYPE; ACOUNT NUMBER; SSAL EMP. SAL%TYPE; BEGIN LOOP EXIT WHEN DNO > 50; SELECT DNAME INTO DPNAME FROM DEPT WHERE DEPTNO=DNO; SELECT AVG(SAL), COUNT(*), SUM(SAL) INTO ASAL, ACOUNT,SSAL FROM EMP WHERE DEPTNO = DNO; DBMS_OUTPUT.PUT_LINE( DEPARTMENT NAME DPNAME); DBMS_OUTPUT.PUT_LINE( DEPARTMENT COUNT ACOUNT); DBMS_OUTPUT.PUT_LINE( AVERAGE SALARY ASAL); DBMS_OUTPUT.PUT_LINE( SUM SALARY SSAL); DNO := DNO+10; END LOOP; END;


P34. WRITE A PL/SQL SCRIPT TO COUNT THE NO: OF EMPLOYEES. IF TOTAL


ENO := &ENO; UPDATE EMP SET SAL =SAL+SAL*5/100 WHERE SAL>4000 AND EMPNO=ENO; END; P38. GIVEN A TABLE TEMP ( SNO NUMBER(3) , DNO NUMBER,TEXT CHAR (4) . WRITE A PL/SQL SCRIPT TO INSERT 10 RECORDS IN TABLE TEMP AS PER THE FOLLOWING SPECIFICA TIONS: SNO DNO TEXT 1 50 ODD 2 100 EVEN 3 150 ODD DECLARE SNO NUMBER := 1; DNO NUMBER ; TEXT CHAR(5) ; BEGIN LOOP EXIT WHEN SNO > 10; DNO := SNO * 50; IF MOD ( SNO, 2 ) = 0 TEXT := EVEN; ELSE TEXT := ODD; INSERT INTO TEMP VALUES (SNO,DNO,TEXT); SNO := SNO + 1; END LOOP; COMMIT; END; P39 . Write PL/SQL code to DELETE the record of an employee whose salary is more than 4000 . Get the empno from the user. DECLARE ENO NUMBER(4); BEGIN ENO := &ENO; DELETE FROM EMP WHERE SAL>4000 AND EMPNO=ENO; END; P40 .Write PL/SQL code to insert a new record in the table emp after obtaining values ( empno, ename, hiredate, sal ) from user. declare tempno number(4); tename varchar(10); thiredate varchar2(12); tsal number(7,2); begin tempno :=&tempno; tename :='&tename'; thiredate :='&thiredate'; tsal := &tsal; insert into emp (empno,ename,hiredate,sal) values(tempno,tename,to_date(thir


edate,'DD-mon-yyyy'),tsal); end; Enter value for tempno: 8000 old 7: tempno:=&tempno; new 7: tempno:=8000; Enter value for tename: mathew old 8: tename :='&tename'; new 8: tename :='mathew'; Enter value for thiredate: 10-jan-2004 old 9: thiredate :='&thiredate'; new 9: thiredate :='10-jan-2004'; Enter value for tsal: 8000 old 10: tsal := &tsal; new 10: tsal := 8000; PL/SQL procedure successfully completed. P41. WRITE A PROGRAM TO CREATE A EMP %ROWTYPE RECORD .ACCEPT THE EMPNO FROM THE USER, AND DISPLAY ALL THE INFORMATION ABOUT THE EMPLOYEE. declare erec emp%rowtype; eno emp.empno%type; begin eno := &eno; select * into erec from emp where empno=eno; dbms_output.put_line( 'Emp no : ' erec.empno) ; dbms_output.put_line( 'Name : ' erec.ename) ; dbms_output.put_line( 'Salary : ' erec.sal); dbms_output.put_line( 'Deptno : ' erec.deptno); dbms_output.put_line( 'hiredate: ' erec.hiredate); dbms_output.put_line( 'job : ' erec.job); end; SQL> / Enter value for eno: 7788 old 5: eno := &eno; new 5: eno := 7788; Emp no : 7788 Name : SCOTT Salary : 1452 Deptno : 20 hiredate: 19-APR-87 job : ANALYST PL/SQL procedure successfully completed. PROGRAM BASED ON EXCEPTION P42. Write PL/SQL script that traps an invalid data type value given and displays a custom error message. DECLARE ENO EMP. EMPNO%TYPE; SNO VARCHAR2(5); NAME EMP.ENMAE%TYPE; BEGIN SNO := &SNO ;


SELECT EMPNO,ENAME INTO ENO,NAME FROM EMP WHERE EMPNO=SNO; DBMS_OUTPUT.PUT_LINE ( NAME ENAME EMPNO EMPNO); EXCEPTION WHEN INVALID_NUMBER THEN DBMS_OUTPUT.PUT_LINE(SNO IS INVALID DATA FOR EMPLOYEE ID); END; PROGRAM BASED ON FUNCTIONS P43 . write a function to create factorial of a number. CREATE OR REPLACE FUNCTION FACT (N NUMBER) RETURN NUMBER IS I NUMBER(10); F NUMBER :=1; BEGIN FOR I IN 1.. N LOOP F:= F*I; END LOOP; RETURN F; END; SQL> / Function created. SQL> select fact(8) from dual; FACT(8) --------- 40320 PROGRAM BASED ON PROCEDURES P44. Write a Procedure to increase the salary for all the employees in the EMP table : SQL> select * from emp; EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO --------- ---------- --------- --------- --------- --------- --------- ----- ---- 7369 SMITH CLERK 7902 17-DEC-80 800 20 7499 ALLEN SALESMAN 7698 20-FEB-81 1600 300 30 14 rows selected. SQL> create or replace procedure inc(i number) is begin update emp set sal =sal+i; end; / Procedure created. To execute the procedures in the PL/SQL block: SQL> declare begin inc(100); end; / PL/SQL procedure successfully completed. SQL> select * from emp; EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO


7369 SMITH CLERK 7902 17-DEC-80 800 20 7499 ALLEN SALESMAN 7698 20-FEB-81 1700 300 30 14 rows selected. P45. WRITE A PROCEDURE TO INCREASE THE SALARY FOR THE SPECIFIED EMPLOLEE USING EMPNO IN THE EMP TABLE BASED ON THE FOLLOWING CRITERIA: INCREASE THE SALARY BY 5% FOR CLERKS, 7% FOR SALESMAN , 10% FOR ANALYST, 20 % FOR MANAGER and 25% FOR PRESIDENT. ACTIVATE USING PL/SQL BLOCK. SQL> SELECT * FROM EMP; EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPT 7369 SMITH CLERK 7902 17-DEC-80 800 20 CREATE OR REPLACE PROCEDURE DESIGNATION(ENO NUMBER) IS BEGIN UPDATE EMP SET SAL=SAL+SAL*5/100 WHERE JOB ='CLERK' AND EMPNO=ENO; UPDATE EMP SET SAL=SAL+SAL*7/100 WHERE JOB='SALESMAN' AND EMPNO=ENO; UPDATE EMP SET SAL=SAL+SAL*10/100 WHERE JOB='ANALYST' AND EMPNO=ENO; UPDATE EMP SET SAL=SAL+SAL*20/100 WHERE JOB='MANAGER' AND EMPNO=ENO; UPDATE EMP SET SAL=SAL+SAL*25/100 WHERE JOB='PRESIDENT' AND EMPNO=ENO; END; SQL> / Procedure created. SQL> DECLARE 2 BEGIN 3 DESIGNATION(7369); 4 END; 5 / PL/SQL procedure successfully completed. SQL> SELECT * FROM EMP; EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPT 7369 SMITH CLERK 7902 17-DEC-80 840 20__

plsqlprograms

Documents

Transcript of plsqlprograms