Sql query related to 1st and 2nd highest salary

Question

Can somebody tell me how this sql query evaluates :

SELECT e.sal 
FROM employees e 
WHERE 1 >= (
    SELECT count(*) 
    FROM employess s 
    WHERE s.sal>e.sal
)

I know this query returns 1st and 2nd highest salary. But I want to understand the working of this query. I want to know how it returns 1st and 2nd highest salary. Please explain in detail.

Answer 1

The subselect

 (SELECT count(*) FROM employess s WHERE s.sal>e.sal)

returns number of employees with salary greater than the current employee you looking at in the top level select .

So the query basically says give me all employees such that have 1 or less other employees earn more than them

Pretty straight forward though horribly inefficient. The easier way would be simply this:

select top 2 * from employees order by sal desc

EDIT: This will return top 2 employees with highest salary, which as Gordon noted is not the same as top 2 distinct salary values, since people may have exactly the same salary. If that is a concern that you'd have to dedupe it first like this:

select top 2 sal from employees group by sal order by sal desc

Answer 2

To get your 1st and 2nd highest salary, I'd use ROW_NUMBER() . That should make it easy to understand the rankings of the salaries. Or if you need to account for ties, look into using RANK() .

Something like this should work:

SELECT sal, Row 
FROM (
   SELECT sal, ROW_NUMBER() OVER(ORDER BY sal DESC) AS Row
   FROM employees e ) t
WHERE Row IN (1,2);

And here is the SQL Fiddle .

Good luck.

Answer 3

EDIT: I missed the e and the s. So my answer is incorrect.

That query should return everything.

(SELECT count(*) FROM employess s WHERE s.sal>e.sal)

Since e.sal will NEVER be > e.sal (1 can never be > 1), thus, the count(*) will always be 0. and SELECT e.sal FROM employees e WHERE 1 >= (0) will always be everything

If you want to return the top 2 salary, you might want to try this:

SELECT TOP 2 * FROM employess AS e ORDER BY e.sal DESC