i want to find all the record against second highest salary in table and there many employee with second highest salary how to do that

Question

I want to find all the records with the second highest salary in the table. There are many such employees, how do I do that?

Table: Employee

ID    salary      emp_name   emp_address                           
1     400         A          abc
2     800         B          def 
3     300         C          hjs
4     400         D          teuu
5     400         E          kakn
6     400         E          kssj

Answer 1

You can use the dense_rank as follows:

Select * from
(Select t.*,
       dense_rank() over (order by salary desc) rn
  From your_table t) t
Where rn = 2

If you are on lower version of mysql in which windows functions are not allowed then you can use corelated query as follows:

Select t.*
  From your_table t
 Where 1 = (select count(distinct tt.salary)
             From your_table tt
            Where tt.salary > t.salary)

Answer 2

You might find it simple enough to do:

select t.*
from t
where t.salary = (select distinct t2.salary
                  from t t2
                  order by t2.salary desc
                  limit 1 offset 1
                 );