How to fetch the nth highest salary from a table without using TOP and sub-query?

Try a CTE - Common Table Expression:

WITH Salaries AS(    SELECT        SalaryAmount, ROW_NUMBER() OVER(ORDER BY SalaryAmount DESC) AS 'RowNum'    FROM        dbo.SalaryTable)SELECT  SalaryAmountFROM  SalariesWHERE   RowNum <= 5

This gets the top 5 salaries in descending order - you can play with the RowNumn value and basically retrieve any slice from the list of salaries.

There are other ranking functions available in SQL Server that can be used, too - e.g. there's NTILE which will split your results into n groups of equal size (as closely as possible), so you could e.g. create 10 groups like this:

WITH Salaries AS(    SELECT        SalaryAmount, NTILE(10) OVER(ORDER BY SalaryAmount DESC) AS 'NTile'    FROM        dbo.SalaryTable)SELECT  SalaryAmountFROM  SalariesWHERE   NTile = 1

This will split your salaries into 10 groups of equal size - and the one with NTile=1 is the "TOP 10%" group of salaries.

;with cte as(Select salary,row_number() over (order by salary desc) as rnfrom salaries)select salary from cte where rn=@n

(or use dense_rank in place of row_number if you want the nth highest distinct salary amount)

Select  * From    Employee E1 WhereN = (Select Count(Distinct(E2.Salary)) From Employee E2 Where E2.Salary >= E1.Salary)