What is query to Find Second Highest Salary Of Employee?

Select distinct Salary from Employee e1 where 2=Select count(distinct Salary) from Employee e2 where e1.salary<=e2.salary;

Alternative Solution : Suggested by Ankit Srivastava

select min(salary)from(select distinct salary from emp order by salary desc)where rownum<=2;

1.Distinct keyword:–>The Distinct keyword is very useful keyword which will eliminate duplicates and fetches the Result.

2.Aliases–> Alias is nothing but the obfuscation technique to protect the real names of database fields

3.Count()–>Count() function is used to count the records in the table.

Step 1: So to understand the above query we will start with simple select statement.

Output:

Employee_num	Employee_name	Department	Salary
1	Amit	OBIEE	680000
2	Rohan	OBIEE	550000
3	Rohit	OBIEE	430000

Here our intention is to fetch the record where salary is second highest.

Step 2 :We will fetch the distinct Salary of employee and give the alias to it.

Select distinct Salary from Employee e1;

Output:

Salary

680000

550000

430000

Step 3 :We need to calculate Second highest salary.So we need to get the count of all distinct salaries.

Select count(distinct Salary) from Employee e2;

Output: 

 3

Step 4 :We actually need to calculate second highest salary.So we will modify the Step 2 query and Step3 Query:

Select distinct Salary from Employee e1 where 2=

Select count(distinct Salary) from Employee e2 where e1.salary<=e2.salary;

Output:

550000

The above query will give us the 2nd highest salary of employee.Hope you will understand the logic behind fetching the second highest salary.There are other ways to find 2nd highest salary of employee using rank and dense_rank function.

Other ways to find second highest salary:

Query 1:

SELECT max(e1.sal), e1.deptno FROM s_emp e1 WHERE sal < (SELECT max(sal) FROM s_emp e2 WHERE e2.deptno = e1.deptno) GROUP BY e1.deptno;

Query 2:

SELECT * FROM (SELECT S.*,DENSE_RANK() OVER (PARTITION BY DNO ORDER BY SALARY DESC) DR FROM SOURCE ) S WHERE S.DR=2;

Query 3:

Select Name From Employees Where Salary =

(Select Distinct Top(1) Salary from Employees where Salary Not In

(Select Dustinct Top(1) Salary from Employees Order By Salary Descending) Order by Salary Descending);

There are some other ways of calculating the second highest salary in different DBMS i.e. Oracle,Mysql,Postgresql:

Type 1 : Using correlated subquery:

SELECT name, salary FROM Employee e1 WHERE N-1 = (SELECT COUNT(DISTINCT salary) FROM Employee e2 WHERE e2.salary > e1.salary)SELECT name, salary FROM Employee e1 WHERE 2-1 = (SELECT COUNT(DISTINCT salary) FROM #Employee e2 WHERE e2.salary > e1.salary);

Using this method of finding out the second highest salary is slow because of the execution of inner query again and again.

Type 2: Microsoft SQL Server (Using Top Function)

SELECT TOP 1 salary FROM ( SELECT DISTINCT TOP N salary FROM Employee ORDER BY salary DESC ) AS temp ORDER BY salary;

Type3 : Mysql (using limit Clause)

SELECT salary FROM Employee ORDER BY salary DESC LIMIT 1,1;

Type 4: Oracle Function (using rownum function)

SELECT Salary FROM ( SELECT e.Salary, ROW_NUMBER() OVER (ORDER BY salary DESC) rn FROM Employee e ) WHERE rn = 2;

Type 5 : One of the website reader has suggested this solution Solution of Gautam Roy :

select * from OrderDetails a where a.Quantity < (select max(b.Quantity) from OrderDetails b) order by a.Quantity DESC limit 1;

Type 6 : Another Reader suggested solution of Bharath chary Vadla

name       sal
———-  ———————
x             120.00
y             140.00
z             200.00
p            400.00

select max(sal) as maxsalary from salary where sal < ( select max(sal) as maxsalary from salary );

Output :

maxsalary

———————

200.00

You can try this also for getting results easy

Type 7 : Suggested by one of the Reader Selva Saleen :

select salary from (select rownum row_num,salary from (select * from EMPLOYEE E order by E.salary desc))where row_num=2;

Type 8 : Find Nth Highest Salary using Limit Keyword Suggested by Reader named Dinesh Singh :

Select name, salary from employee order by salary desc limit(1,n-1);

Type 9 : New Solution  suggested by reader named Victor :

select distinct e2.salary
from (select A.Employee_num , A.salary , count(distinct B.salary) as salary_distinct_rank
from Employee A
left join
Employ B
where B.salary <= A.salary
group by A.Employee_num, A.salary ) e2

where e2.salary_distinct_rank = 2 ;