pl/sql - using if statements please use speaker notes for additional information!

PL/SQL - Using IF statements

Please use speaker notes for additional information!

PL/SQL - IFPL/SQL - IF

SQL> CREATE TABLE first_pay_new 2 AS 3 SELECT * FROM first_pay;

Table created.

SQL> SELECT * FROM first_pay_new;

PAY_ NAME JO STARTDATE SALARY BONUS---- -------------------- -- --------- --------- ---------1111 Linda Costa CI 15-JAN-97 45000 10002222 John Davidson IN 25-SEP-92 40000 15003333 Susan Ash AP 05-FEB-00 25000 5004444 Stephen York CM 03-JUL-97 42000 20005555 Richard Jones CI 30-OCT-92 50000 20006666 Joanne Brown IN 18-AUG-94 48000 20007777 Donald Brown CI 05-NOV-99 450008888 Paula Adams IN 12-DEC-98 45000 2000

8 rows selected.

Step 1 for me was to create a new table to use since I will be doing some updating in these examples.


SQL> edit firstpay1

SET VERIFY OFFDECLARE v_pay_id first_pay_new.pay_id%TYPE :=&input_pay_id; v_new_sal first_pay_new.salary%TYPE; v_jobcode first_pay_new.jobcode%TYPE;BEGIN SELECT salary, jobcode INTO v_new_sal, v_jobcode FROM first_pay_new WHERE pay_id = v_pay_id; IF v_jobcode = 'CI' THEN v_new_sal := v_new_sal * 1.05; END IF; UPDATE first_pay_new SET salary = v_new_sal WHERE pay_id = v_pay_id;END;/SET VERIFY ON

Declare the variables and assign &input_pay_id to v_pay_id so that it will require user input.

SELECT one record for a given person and store the jobcode and salary into the specified variables.

IF the jobcode on the selected record = CI then the salary stored in the variable is increased by 5% and stored back in the variable using the assignment sign.

NOTE the syntax of IF condition THEN processing;END IF;

The update changes the salary by making it equal to the variable for the specified record.

PL/SQL - IFPL/SQL - IF SQL> SELECT * FROM first_pay_new;


SQL> @firstpay1Enter value for input_pay_id: 5555

PL/SQL procedure successfully completed.



This is the output from the previous slide. The user enters 5555 as the pay_id and since Richard Jones has a jobcode of CI, he receives a 5% raise.




SQL> @ firstpay1Enter value for input_pay_id: 3333




The record 3333 does not have a jobcode of CI so no change was made to the salary.


SQL> edit firstpay1a

SET VERIFY OFFDECLARE v_pay_id first_pay_new.pay_id%TYPE :=&input_pay_id; v_new_sal first_pay_new.salary%TYPE; v_jobcode first_pay_new.jobcode%TYPE;BEGIN SELECT salary, jobcode INTO v_new_sal, v_jobcode FROM first_pay_new WHERE pay_id = v_pay_id; IF v_jobcode = 'CI' THEN v_new_sal := v_new_sal * 1.05; UPDATE first_pay_new SET salary = v_new_sal WHERE pay_id = v_pay_id; END IF;END;/SET VERIFY ON

The difference in this code is that the UPDATE is now processing within the IF. The update will only be done IF the jobcode that came in with the record was CI.



SQL> @ firstpay1aEnter value for input_pay_id: 1111





IF v_jobcode= ‘CI’

Y

Calculate new salary

Update record on table

N

IF v_jobcode = 'CI' THEN v_new_sal := v_new_sal * 1.05; UPDATE first_pay_new SET salary = v_new_sal WHERE pay_id = v_pay_id;END IF;


SET VERIFY OFFDECLARE v_pay_id first_pay_new.pay_id%TYPE :=&input_pay_id; v_new_sal first_pay_new.salary%TYPE; v_jobcode first_pay_new.jobcode%TYPE;BEGIN SELECT salary, jobcode INTO v_new_sal, v_jobcode FROM first_pay_new WHERE pay_id = v_pay_id; IF v_jobcode = 'CI' THEN v_new_sal := v_new_sal * 1.05; ELSE v_new_sal := v_new_sal * 1.02; END IF; UPDATE first_pay_new SET salary = v_new_sal WHERE pay_id = v_pay_id;END;/SET VERIFY ON

SQL> edit firstpay2

If the jobcode stored in the variable is CI then the salary will be increased by 5% otherwise the salary will be increased by 2%





8 rows selected.




Since jobcode is not CI, the salary is increased by 2%.



Y


with 5% increase

N


with 2% increase

IF v_jobcode = 'CI' THEN v_new_sal := v_new_sal * 1.05; ELSE v_new_sal := v_new_sal * 1.02; END IF;


SET VERIFY OFFDECLARE v_pay_id first_pay_new.pay_id%TYPE :=&input_pay_id; v_new_sal first_pay_new.salary%TYPE; v_jobcode first_pay_new.jobcode%TYPE;BEGIN SELECT salary, jobcode INTO v_new_sal, v_jobcode FROM first_pay_new WHERE pay_id = v_pay_id; IF v_jobcode = 'CI' THEN v_new_sal := v_new_sal * 1.05; ELSE IF v_jobcode = 'IN' THEN v_new_sal := v_new_sal * 1.03; ELSE v_new_sal := v_new_sal * 1.01; END IF; END IF; UPDATE first_pay_new SET salary = v_new_sal WHERE pay_id = v_pay_id;END;/SET VERIFY ON

SQL> edit firstpay3

The first condition is testing to see if the v_jobcode = CI. If it does, then a 5% increase is calculated. If it is not true, then the false processing after the ELSE asks another question (an inner if). If the inner if condition which is if v_jobcode = IN is true then 1 3% increase is calculated. If the inner if is false then the ELSE condition is executed which calculates a 1% increase.

Note that each IF has an END IF.








Since the jobcode is IN, the salary is increased by 3%.








8 rows selected.

The conditions are both false so the else to the second/inner if is executed and salary is increased by 1%.


Y


with 5% increase

N

IF v_jobcode = 'CI' THEN v_new_sal := v_new_sal * 1.05; ELSE IF v_jobcode = 'IN' THEN v_new_sal := v_new_sal * 1.03; ELSE v_new_sal := v_new_sal * 1.01; END IF; END IF;


with 3% increase

IF v_jobcode= ‘IN’

Y

N


with 3% increase


SET VERIFY OFFDECLARE v_pay_id first_pay_new.pay_id%TYPE :=&input_pay_id; v_new_sal first_pay_new.salary%TYPE; v_jobcode first_pay_new.jobcode%TYPE;BEGIN SELECT salary, jobcode INTO v_new_sal, v_jobcode FROM first_pay_new WHERE pay_id = v_pay_id; IF v_jobcode = 'CI' THEN v_new_sal := v_new_sal * 1.05; ELSIF v_jobcode = 'IN' THEN v_new_sal := v_new_sal * 1.03; ELSE v_new_sal := v_new_sal * 1.01; END IF; UPDATE first_pay_new SET salary = v_new_sal WHERE pay_id = v_pay_id;END;/SET VERIFY ON

SQL> edit firstpay4









Processing went to the ELSE where a 1% increase was given.








Record 6666 has a jobcode of IN so the answer to the ELSIF was true and the salary was increased by 3%.

pl/sql - using if statements please use speaker notes for additional information!

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