To find the Nth Highest Salary - sql

Question

Write a SQL query to get the nth highest salary from the Employee table (SQL Server)
| Id | Salary |
+----+--------+
| 1 | 100 |
| 2 | 200 |
| 3 | 300 |
For this example, the nth highest salary where n = 2 is 200. If there is no nth highest salary, then the query should return null.
| getNthHighestSalary(2) |
+------------------------+
| 200 |
Is there any way to write this query other than by using function?

Answer 1

Another possible approach is using ROW_NUMBER() or DENSE_RANK() functions. It is important to know if there can be more then one salary at or before Nth position.
CREATE TABLE #Salary (
[id] int,
[salary] numeric(10, 2)
)
INSERT #Salary
([id], [salary])
VALUES
(1, 100),
(2, 500),
(3, 200),
(4, 300),
(5, 200);
DECLARE #Position int = 2;
-- With possible duplicate salaries
WITH cte AS (
SELECT
[id], [salary],
DENSE_RANK() OVER (ORDER BY [salary] ASC) AS [DRPosition]
FROM #Salary
)
SELECT [id]
FROM cte
WHERE [DRPosition] = #Position
ORDER BY [DRPosition];
-- Without possible duplicate salaries
WITH cte AS (
SELECT
[id], [salary],
ROW_NUMBER() OVER (ORDER BY [salary] ASC) AS [RPosition]
FROM #Salary
)
SELECT [id]
FROM cte
WHERE [RPosition] = #Position
ORDER BY [RPosition]

Answer 2

The following does almost exactly what you want:
select salary
from employee
order by salary desc
offset <n> rows fetch next 1 row only;
The only problem is that it does not return NULL when there is no such salary. You can handle using a subquery:
select (select salary
from employee
order by salary desc
offset <n> rows fetch next 1 row only
) as salary;
If you want ties to have the same ranking, then use select distinct salary in the subquery.

Answer 3

You can try this for getting n-th highest salary, where n = 1,2,3....(int)
SELECT TOP 1 salary FROM (
SELECT TOP n salary
FROM employees
ORDER BY salary DESC) AS emp
ORDER BY salary ASC
Hope this will help you. Below is one of the implementation.
create table #salary (salary int)
insert into #salary values (100), (200), (300), (400), (500)
SELECT TOP 1 salary FROM (
SELECT TOP 3 salary
FROM #salary
ORDER BY salary DESC) AS emp
ORDER BY salary ASC
drop table #salary
The output is here 300 as 500 is first highest, 400 is second highest and 300 is the third highest as shown below
salary
300
Here n is 3

Answer 4

177. Nth Highest Salary - Leetcode
Try this out ! It worked !
CREATE FUNCTION getNthHighestSalary(N INT) RETURNS INT
BEGIN
SET N = N-1;
RETURN (
SELECT DISTINCT Salary as "getNthHighestSalary(2)"
FROM Employee
ORDER BY Salary DESC
LIMIT 1 OFFSET N
);
END

Answer 5

You can try this using row_number()
WITH CTE AS
(
SELECT EmpID, Salary,
RN = ROW_NUMBER() OVER (ORDER BY Salary DESC)
FROM Employee
)
SELECT EmpID, Salary
FROM CTE
WHERE RN = n

Answer 6

Finds the salary for which there exist exactly 8 higher salaries,
meaning finds the 9th highest salary.
Returns NULL if there is no 9th Salary:
SELECT DISTINCT e.Salary FROM Employee e WHERE
(SELECT COUNT(*) FROM Employee ie WHERE ie.Salary > e.Salary) = 8
UNION
SELECT NULL WHERE (SELECT COUNT(Salary) FROM Employee) < 9

Answer 7

You can simply try this using correlated sub-query to find Nth highest salary in Employee table..
SELECT *
FROM Employee Emp1
WHERE (N-1) = (
SELECT COUNT(DISTINCT(Emp2.Salary))
FROM Employee Emp2
WHERE Emp2.Salary > Emp1.Salary)

Answer 8

I tried this query and it worked
select sal from (select sal from employee
order by sal desc
limit n)
order by sal asc
limit 1

Answer 9

SELECT DISTINCT(Salary) FROM Employee ORDER BY Salary DESC LIMIT n-1,1;

Answer 10

The best approach
select salary from emp as e1
where N-1 (select count(distinct salary) from emp as e2
where e2.salary > e1.salary);

Answer 11

Another Possible approach.I think this could help .
select min(T.Salary) from
(
select distinct top n Salary from Employee order by Salary desc
)T