FINDING NTH HIGHEST NUMBER

EmpID   Name    Salary
1   A   100
2   B   800
3   C   300
4   D   400
5   E   500
6   F   200
7   G   600

SELECT * FROM Employee E1
WHERE (N-1) = (
                SELECT COUNT(DISTINCT(E2.Salary))
                FROM Employee E2
                WHERE E2.Salary > E1.Salary
              )

Explanation

Suppose you want to find 5th highest salary, which means there are total 4 employees who have salary greater than 5th highest employee. So for each row from the outer query check the total number of salaries which are greater than current salary. Outer query will work for 100 first and check for number of salaries greater than 100. It will be 6, do not match (5-1) = 6 where clause of outerquery. Then for 800, and check for number of salaries greater than 800, 4=0 false then work for 300 and finally there are totally 4 records in the table which are greater than 300. Therefore 4=4 will meet the where clause and will return 3 C 300.

Another way

SELECT *
FROM(
SELECT BONUSMONTHID, dense_rank() over (ORDER BY BONUSMONTHID DESC) AS DenseRank
FROM REFERRALBONUS )
where DenseRank = 2