Second Highest Salary

Question

Write a SQL query to get the second highest salary from the Employee table.

    | Id | Salary |
    | 1  | 100    |
    | 2  | 200    |
    | 3  | 300    |
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Answer 1

Query:

CREATE TABLE a
    ([Id] int, [Salary] int)
;
     
INSERT INTO a
    ([Id], [Salary])
VALUES
    (1, 100),
    (2, 200),
    (3, 300)
;

GO
SELECT Salary as SecondHighestSalary
FROM a 
ORDER BY Salary 
OFFSET 1 ROWS FETCH NEXT 1 ROWS ONLY

| SecondHighestSalary |
| ------------------: |
|                 200 |

Answer 2

You can use RANK() function to rank the values for Salary column.

SELECT *
FROM
(
 SELECT *, RANK()OVER(ORDER BY Salary DESC) As SalaryRank
 FROM Employee 
) AS Tab
WHERE SalaryRank = 2

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)

SELECT TOP 1 salary FROM (
   SELECT TOP 2 salary 
   FROM #salary 
   ORDER BY salary DESC) AS emp 
ORDER BY salary ASC

drop table #salary

The output is here 200 as 300 is first highest, 200 is second highest and 100 is the third highest as shown below

salary
200

Here n is 2

Answer 4

While you can use a CTE (from MSSQL 2005 or newer) or ROWNUMBER the easiest and more "portable" way is to just order by twice using a subquery.

select top 1 x.* from
(select top 2 t1.* from dbo.Employee t1 order by t1.Salary) as x
order by x.Salary desc

The requisite to show null when there's not a second bigger salary is a bit more tricky but also easy to do with a if.

if (select count(*) from dbo.Employee) > 1
begin
    select top 1 x.* from
    (select top 2 emp.* from dbo.Employee emp order by emp.Salary) as x
    order by x.Salary desc
end
else begin
    select null as Id, null as Salary
end

Obs:. OP don't said what to do when the second largest is a tie with the first but using this solution is a simple matter of using a DISTINCT in the IF subquery.

Answer 5

I would use DENSE_RANK() & do LEFT JOIN with employee table :

SELECT t.Seq, e.*
FROM ( VALUES (2) 
     ) t (Seq) LEFT JOIN
     (SELECT e.*,
             DENSE_RANK() OVER (ORDER BY Salary DESC) AS Num
      FROM Employee e
     ) e 
     ON e.Num = t.Seq;

Answer 6

In case of ties you want the second highest distinct value. E.g. for values 100, 200, 300, 300, you want 200.

So get the highest value (MAX(salary) => 300) and then get the highest value less than that:

select max(salary) from mytable where salary < (select max(salary) from mytable);