List all employees having percentage greater than some number

Question

How to list all employees having percentage greater than some number, and percentage gets calculated on the basis of max salary.

Example:

if max salary is 100 and an employee salary is 50 then percentage should show 50.

This is what i tried:

select (salary/Max(salary)*100) as percentage from test.employeetable
where percentage > 75;

Error that i get is:

Unknown column 'percentage' in 'where clause

Answer 1

Try something like that;

select * from (
select (salary / (select max(salary) from test.employeetable) * 100) as percentage from test.employeetable) Records
where percentage > 75;

Answer 2

First get the max in a variable, then select the relevant results

select @max := max(salary) from test.employeetable;
select (salary/@max*100) as pc from test.employeetable having pc > 75;

note the having instead of where .

You could display the relevant salary (etc...) as well

select @max := max(salary) from test.employeetable;
select salary,(salary/@max*100) as pc from test.employeetable having pc > 75;

Without using a variable

select (salary/m.mx*100) as pc from test.employeetable, 
(select max(salary) as mx from test.employeetable) as m
having pc > 75;

Answer 3

Select (salary/Max(salary)*100) as percentage 
from test.employeetable 
where (salary/Max(salary)*100) > 75;  

in MySql you Can not use an "as" in where clause

Answer 4

Use a subquery

SELECT *
FROM (
  SELECT (salary / MAX(salary) * 100) AS percentage
  FROM test.employeetable
) x
WHERE percentage > 75