What will be the SQL query to select the second highest salary from the employee table?

You can use Rank() analytical function to achieve this.

the2 nd

SELECT max(salary) FROM Employee WHERE salary NOT IN (SELECT max(salary) FROM Employee)

OR

SELECT FROM Employee E1 WHERE (1) = SELECT COUNT(DISTINCT(E2.Salary)) FROM Employee E2 WHERE E2.Salary > E1.Salary)

the N th

SELECT FROM Employee E1 WHERE (N-1) = SELECT COUNT(DISTINCT(E2.Salary)) FROM Employee E2 WHERE E2.Salary > E1.Salary)

select  max(salary)

from     employees

where  salary <  (select max(salary) from employees)

SELECT max(salary) FROM Employee WHERE salary NOT IN (SELECT max(salary) FROM Employee);

select level,max(sal) from emp

where level =2

connect by prior (sal) > sal

group by level

select Max(Sal) from emp where Sal < (select Max(Sal) from emp)

Select salary from

( select salary from employees order by salary desc )

Where rownum=2;

 

Please notice that rownum is pseudo column

select distinct sal from emp where sal = (select max(e.sal) from emp ewhere sal < (select max(sal) from emp ) ) order  by sal desc

select SALARY from EMPLOYEE EMP

where

2=(

select count(distinct SALARY) from EMPLOYEE

where

EMP.SALARY<=SALARY

);

 

This would work. You may give it a try...

Also you may replace2 with any number n to find nth highest salary.

<p>select E1.* from employees E inner join ( select E.empid,e.Salary, dens_rank over (order by Salary desc) as RNk from employee E ) T on E1.empid = T.empid where T.RNK =2</p>

SELECT * FROM hr.employees WHERE salary = (SELECT salary FROM (SELECT salary, ROW_NUMBER () OVER (ORDER BY salary DESC) r FROM hr.employees GROUP BY salary) WHERE r = :salary_rank)