/* sql 2nd highest salary employee */
select sal, ename
from emp
where sal =
(
select max(sal) from emp where sal <
(select max(sal) from emp)
)
----------------------------------------------- option 2
select *
from
(
select ename, sal, dense_rank() over(order by sal desc) rank
from emp
)
where rank =2;select *from employee
group by salary
order by salary desc limit 1,1;Here is the solution for nth highest
salary from employees table
SELECT FIRST_NAME , SALARY FROM
(SELECT FIRST_NAME, SALARY, DENSE_RANK() OVER
(ORDER BY SALARY DESC) AS SALARY_RANK
FROM EMPLOYEES)
WHERE SALARY_RANK = n; SELECT MAX(SALARY) 'SECOND_MAX' FROM EMPLOYEES
WHERE SALARY <> (SELECT MAX(SALARY) FROM EMPLOYEES);/* Highest salary by Department/Location */
SELECT e.ename, e.sal, e.deptno, d.loc
FROM emp e
JOIN dept d
ON e.deptno = d.deptno
WHERE e.sal in
(
select max(sal)
from emp
group by deptno
)