How to get second-highest salary employees in a table

Question

It\'s a question I got this afternoon:

There a table contains ID, Name, and Salary of Employees, get names of the second-highest salary employees, in SQL Server

Answer 1

I think you would want to use DENSE_RANK as you don't know how many employees have the same salary and you did say you wanted nameS of employees.

CREATE TABLE #Test
(
    Id INT,
    Name NVARCHAR(12),
    Salary MONEY
)

SELECT x.Name, x.Salary
FROM
        (
        SELECT  Name, Salary, DENSE_RANK() OVER (ORDER BY Salary DESC) as Rnk
        FROM    #Test
        ) x
WHERE x.Rnk = 2

ROW_NUMBER would give you unique numbering even if the salaries tied, and plain RANK would not give you a '2' as a rank if you had multiple people tying for highest salary. I've corrected this as DENSE_RANK does the best job for this.

Answer 2

Try this to get the respective nth highest salary.

SELECT
    *
FROM
    emp e1
WHERE
    2 = (
        SELECT
            COUNT(salary)
        FROM
            emp e2
        WHERE
            e2.salary >= e1.salary
    )

Answer 3

Simple way WITHOUT using any special feature specific to Oracle, MySQL etc.

Suppose EMPLOYEE table has data as below. Salaries can be repeated.

By manual analysis we can decide ranks as follows :-

Same result can be achieved by query

select  *
from  (
select tout.sal, id, (select count(*) +1 from (select distinct(sal) distsal from     
EMPLOYEE ) where  distsal >tout.sal)  as rank  from EMPLOYEE tout
) result
order by rank

First we find out distinct salaries. Then we find out count of distinct salaries greater than each row. This is nothing but the rank of that id. For highest salary, this count will be zero. So '+1' is done to start rank from 1.

Now we can get IDs at Nth rank by adding where clause to above query.

select  *
from  (
select tout.sal, id, (select count(*) +1 from (select distinct(sal) distsal from     
EMPLOYEE ) where  distsal >tout.sal)  as rank  from EMPLOYEE tout
) result
where rank = N;

Answer 4

declare

cntr number :=0;

cursor c1 is

select salary from employees order by salary desc;

z c1%rowtype;

begin

open c1;

fetch c1 into z;

while (c1%found) and (cntr <= 1) loop


cntr := cntr + 1;

fetch c1 into z;

dbms_output.put_line(z.salary);

end loop;

end;

Answer 5

Try this: This will give dynamic results irrespective of no of rows

SELECT * FROM emp WHERE salary = (SELECT max(e1.salary) 
FROM emp e1 WHERE e1.salary < (SELECT Max(e2.salary) FROM emp e2))**

Answer 6

Creating temporary table

Create Table #Employee (Id int identity(1,1), Name varchar(500), Salary int)

Insert data

Insert Into #Employee
    Select 'Abul', 5000
Union ALL 
    Select 'Babul', 6000
Union ALL 
    Select 'Kabul', 7000
Union ALL 
    Select 'Ibul', 8000
Union ALL 
    Select 'Dabul', 9000

Query will be

select top 1 * from #Employee a
Where a.id <> (Select top 1 b.id from #Employee b ORDER BY b.Salary desc)
order by a.Salary desc

Drop table

drop table #Empoyee