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In the one and only true way. The object-oriented version of 'Spaghetti code' is, of course, 'Lasagna code'. (Too many layers). Roberto Waltman

Implement IF-THEN-ELSE logic in a SELECT statement

April 12, 2016 7:26 pm

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Implement IF-THEN-ELSE logic in a SELECT statement



												 desc HR.employees;
												 
												Name           Null     Type         
												-------------- -------- ------------ 
												EMPLOYEE_ID    NOT NULL NUMBER(6)    
												FIRST_NAME              VARCHAR2(20) 
												LAST_NAME      NOT NULL VARCHAR2(25) 
												EMAIL          NOT NULL VARCHAR2(25) 
												PHONE_NUMBER            VARCHAR2(20) 
												HIRE_DATE      NOT NULL DATE         
												JOB_ID         NOT NULL VARCHAR2(10) 
												SALARY                  NUMBER(8,2)  
												COMMISSION_PCT          NUMBER(2,2)  
												MANAGER_ID              NUMBER(6)    
												DEPARTMENT_ID           NUMBER(4)  

												

The following query display salary status based on the salary.


Solution

   					 
												SELECT first_name, last_name, CASE WHEN salary = 6000 THEN 'Minimum wage'
																   WHEN salary >= 24000 THEN 'Over paid'
																   ELSE 'Under paid'
															  END AS "Salary Status"
												FROM   hr.employees;
																    

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