SQL Query to Find Second Highest Salary

There are a lot of real-world scenarios where a query for finding the nth salary will be really useful. Considering finding the second-highest salary in SQL, we have one sample table.

Finding Second Highest Salary

Consider below simple table: 

Name	Salary
Aman	100000
Shubham	1000000
Naveen	40000
Nishant	500000

How to find the employee whose salary is the second highest. For example, in the above table, “Nishant ” has the second highest salary at 500000. 

Below is a simple query to find the employee whose salary is the highest. 

select * from employee where salary=(select Max(salary) from employee);

Note: Depending on the default settings and MySQL version, we may receive ERROR 1140 when running this query on the MySQL database. The solution can be found in the article’s final section. 

We can nest the above SQL query to find the second-largest salary. 

select *from employee 
group by salary 
order by  salary desc limit 1,1;

Output: 

Name	Salary
Nishant	500000

Other Ways to Find Second Highest Salary in SQL

SELECT name, MAX(salary) AS salary 
FROM employee 
WHERE salary IN
(SELECT salary FROM employee MINUS SELECT MAX(salary) 
FROM employee); 

SELECT name, MAX(salary) AS salary 
FROM employee 
WHERE salary <> (SELECT MAX(salary) 
FROM employee);

In SQL Server using Common Table Expression or CTE, we can find the second highest salary: 

WITH T ASd
(
SELECT *
   DENSE_RANK() OVER (ORDER BY Salary Desc) AS Rnk
FROM Employees
)
SELECT Name
FROM T
WHERE Rnk=2;

How to find the third-largest salary? 

Simple, We can do one more nesting. 

SELECT name, MAX(salary) AS salary
FROM employee
WHERE salary < (SELECT MAX(salary) 
                 FROM employee
                 WHERE salary < (SELECT MAX(salary)
                 FROM employee)
                ); 

Note that instead of nesting for the second, third, etc largest salary, we can find the nth salary using a general query like in MySQL:

SELECT salary 
FROM employee 
ORDER BY salary desc limit n-1,1

SELECT name, salary
FROM employee A
WHERE n-1 = (SELECT count(1) 
             FROM employee B 
             WHERE B.salary>A.salary)

If multiple employees have the same salary:

Suppose you have to find 4th highest salary 

SELECT * FROM employee 
WHERE salary= (SELECT DISTINCT(salary) 
FROM employee ORDER BY salary DESC LIMIT 3,1);

The generic query will be 

SELECT * FROM employee 
WHERE salary= (SELECT DISTINCT(salary) 
FROM employee ORDER BY salary DESC LIMIT n-1,1);

Solution for ERROR 1140 :

While querying the database to fetch an employee with maximum salary, we might get ERROR 1140:

mysql> SELECT name, MAX(salary) AS salary FROM employee;
ERROR 1140 (42000): In aggregated query without GROUP BY, expression #1 of SELECT list contains nonaggregated column 
'database.employee.name'; this is incompatible with sql_mode=only_full_group_by

We can clearly understand from the error message that our query is an aggregation query from ‘MAX(salary)’ and at the same time, it uses an unaggregated column ‘name’, which creates ambiguity for MySQL. This error will be captured by those versions of MySQL which contain the value ‘only_full_group_by’ in the ‘sql_mode’ variable. We can check this variable in MySQL using the following command.

SHOW VARIABLES LIKE "sql_mode";

To avoid confusion, we must avoid using the aggregated column and the unaggregated column in the same query to eliminate this error. The following command would help in this.

// EMPLOYEE WITH HIGHEST SALARY
SELECT name, salary FROM employee ORDER BY salary DESC LIMIT 1; 

// EMPLOYEE WITH SECOND HIGHEST SALARY
SELECT name, salary FROM employee WHERE salary < (SELECT MAX(salary) FROM employee) ORDER BY salary DESC LIMIT 1;

// EMPLOYEE WITH Nth HIGHEST SALARY
SELECT name, salary FROM employee ORDER BY salary DESC LIMIT (N-1), 1;

SQL Interview Question – 2nd Highest Salary

This Solution is provided by Mohit.
Kartik contributes to this article. Please write comments if you find anything incorrect or want to share more information about the topic discussed above.