how to get second highest salary department wise without using analytical functions?

Question

Suppose we have 3 employees in each department.we have total 3 departments . Below is the sample source table

Emp deptno salary
A    10     1000
B    10              


        

Answer 1

On MySQL this how you can get second highest salary, given table name is salaries:

By Nested Queries: (where you can change offset 0/1/2 for first, second and third place respectively)

select
  *
from
  salaries as t1 
where 
  t1.salary = (select 
               salary
             from 
               salaries
             where 
             salaries.deptno = t1.deptno ORDER by salary desc limit 1 offset 1);

or might be by creating rank: (where you can change rank= 1/2/3 for first, second and third place respectively)

SET @prev_value = NULL;
SET @rank_count = 0;
select * from 
(SELECT
  s.*, 
  CASE 
      WHEN @prev_value = deptno THEN @rank_count := @rank_count + 1
      WHEN @prev_value := deptno THEN @rank_count := 1 
      ELSE @rank_count := 1 
  END as rank
FROM salaries s
ORDER BY deptno, salary desc) as t
having t.rank = 2;

Answer 2

This will give you 2nd highest salary in each department:

SELECT a.Emp, a.deptno, a.salary
FROM Emp a
WHERE 1 = (SELECT COUNT(DISTINCT salary) 
        FROM Emp b 
        WHERE b.salary > a.salary AND a.deptno = b.deptno)
group by a.deptno

Answer 3

SQL Query:

select TOP 2 max(salary),Emp from EMployee where deptno='your_detpno'

Answer 4

select min(salary),deptno from
(SELECT distinct top 2 salary,deptno from table ORDER BY salary DESC)
as  a 
group by deptno

Answer 5

You can find 2nd highest salary something like this:

select max(a.Salary),a.Deptno from Employee a join (select MAX(salary) salary 
from Employee group by Deptno) b on a.Salary < b.salary group by a.Deptno

And no MAX() is not an analytic function.

Reference

Answer 6

It is a pain, but you can do it. The following query gets the second highest salary:

select t.deptno, max(t.salary) as maxs
from table t
where t.salary < (select max(salary)
                  from table t2
                  where t2.deptno = t.deptno
                 )
group by t.deptno;

You can then use this to get the employee:

select t.*
from table t join
     (select t.deptno, max(t.salary) as maxs
      from table t
      where t.salary < (select max(salary)
                        from table t2
                        where t2.deptno = t.deptno
                       )
      group by t.deptno
     ) tt
     on t.deptno = tt.deptno and t.salary = tt.maxs;