How to find third or nᵗʰ maximum salary from salary table?

Question

How to find third or n^th maximum salary from salary table(EmpID, EmpName, EmpSalary) in optimized way?

Answer 1

Row(s) with the nth highest salary

(assuming that multiple employees can have same salary; i.e. ties can exist)

If you want ALL the rows with the nth highest salary

select * from table_name 
where salary = (
    select distinct(salary) as sal from 
    table_name order by sal desc 
    limit n-1, 1  
);

If you want only FIRST k rows with the nth highest salary

select * from table_name 
where salary = (
    select distinct(salary) as sal from 
    table_name order by sal desc 
    limit n-1, 1  
) limit k;  -- you can also apply `order by` in this line

Reference

• GFG: SQL query to find second highest salary?

Answer 2

Answering this question from the point of view of SQL Server as this is posted in the SQL Server section.

There many approaches of getting Nth salary and we can classify these approaches in two sections one using ANSI SQL approach and other using TSQL approach. You can also check out this find nth highest salary youtube video which shows things practically. Let’s try to cover three ways of writing this SQL.

• Approach number 1: - ANSI SQL: - Using Simple order by and top keyword.
• Approach number 2: - ANSI SQL: - Using Co-related subqueries.
• Approach number 3: - TSQL: - using Fetch Next

Approach number 1: - Using simple order by and top.

In this approach we will using combination of order by and top keyword. We can divide our thinking process in to 4 steps: -

Step 1: - Descending :- Whatever data we have first make it descending by using order by clause.

Step 2:- Then use TOP keyword and select TOP N. Where N stands for which highest salary rank you want.

Step 3: - Ascending: - Make the data ascending.

Step 4:- Select top 1 .There you are done.

So, if you put down the above 4 logical steps in SQL it comes up something as shown below.

Below is the text of SQL in case you want to execute and test the same.

select top 1 * from (select top 2 EmployeeSalary from tblEmployee order by EmployeeSalary desc) as innerquery order by EmployeeSalary asc

Parameterization issue of Approach number 1

One of the biggest issues of Approach number 1 is “PARAMETERIZATION”.

If you want to wrap up the above SQL in to a stored procedure and give input which top salary you want as a parameter, it would be difficult by Approach number 1.

One of the things you can do with Approach number 1 is make it a dynamic SQL but that would not be an elegant solution. Let’s check out Approach number 2 which is an ANSI SQL approach.

Approach number 2: - Using Co-related subqueries.

Below is how co-related subquery solution will look like. In case you are new to Co-related subquery. Co-related subquery is a query which a query inside query. The outer query first evaluates, sends the record to the inner query, inner query then evaluates and sends it to the outer query.

“3” in the query is the top salary we want to find out.

Select E1.EmployeeSalary from tblEmployee as E1 where 3=(Select count(*) from tblEmployee as E2 Where E2.EmployeeSalary>=E1.EmployeeSalary)

So in the above query we have an outer query:-

Select E1.EmployeeSalary from tblEmployee as E1

and inner query is in the where clause. Watch those BOLD’s which indicate how the outer table alias is referred in the where clause which makes co-related evaluate inner and outer query to and fro: -

where 3=(Select count(*) from tblEmployee as E2 Where E2.EmployeeSalary>=E1.EmployeeSalary)

So now let’s say you have records like 3000, 4000 ,1000 and 100 so below will be the steps: -

1. First 3000 will be send to the inner query.
2. Inner query will now check how many record values are greater than or equal to 3000. If the number of record counts is not equal, it will take next value which is 4000. Now for 3000 there are only 2 values which is greater than or equal, 3000 and 4000. So, Is number record count 2>-=3? .NO, so it takes second value which is 4000.
3. Again for 4000 how many record values are greater than or equal. If the number of record count is not equal, it will take next value which is 1000.
4. Now 1000 has 3 records more or equal than 1000, (3000,4000 and 1000 himself). This is where co-related stops and exits and gives the final output.

Approach number 3: - TSQL fetch and Next.

Third approach is by using TSQL. By using Fetch and Next, we can get the Nth highest easily.

But please do note, TSQL code will not work for other databases we will need to rewrite the whole code again.

It would be a three-step process:-

Step 1 Distinct and Order by descending: - First apply distinct and order by which made the salaries descending as well as weed off the duplicates.

Step 2 Use Offset: - Use TSQL Offset and get the top N-1 rows. Where N is the highest salary we want to get. Offset takes the number of rows specified, leaving the other rows. Why (N-1) because it starts from zero.

Step 3 Use Fetch: - Use fetch and get the first row. That row has the highest salary.

The SQL looks something as shown below.

Performance comparison

Below is the SQL plan for performance comparison. Below is the plan for top and order by.

Below is the plan for co-related queries. You can see the number of operators are quiet high in numbers. So surely co-related would perform bad for huge data.

Below is TSQL query plan which is better than cor-related.

So, summing up we can compare more holistically as given in the below table.

Answer 3

select * from employee order by salary desc;

+------+------+------+-----------+
| id   | name | age  | salary    |
+------+------+------+-----------+
|    5 | AJ   |   20 | 100000.00 |
|    4 | Ajay |   25 |  80000.00 |
|    2 | ASM  |   28 |  50000.00 |
|    3 | AM   |   22 |  50000.00 |
|    1 | AJ   |   24 |  30000.00 |
|    6 | Riu  |   20 |  20000.00 |
+------+------+------+-----------+




select distinct salary from employee e1 where (n) = (select count( distinct(salary) ) from employee e2 where e1.salary<=e2.salary);

Replace n with the nth highest salary as number.

Answer 4

In SQL Server 2012+, OFFSET...FETCH would be an efficient way to achieve this:

DECLARE @N AS INT;
SET @N = 3;

SELECT
    EmpSalary
FROM
    dbo.Salary
ORDER BY
    EmpSalary DESC
OFFSET (@N-1) ROWS
FETCH NEXT 1 ROWS ONLY

Answer 5

SELECT TOP 1 salary 
FROM   employee 
WHERE  salary IN (SELECT DISTINCT TOP 3 salary 
                  FROM   employee 
                  ORDER  BY salary DESC) 
ORDER  BY salary ASC 

Answer 6

finding Nth max value using CTE and FIRST_VALUE function. -- 5th max salary

;WITH CTE_NTH_SAL AS 
    (SELECT FIRST_VALUE(ESAL) OVER(ORDER BY ESAL DESC) AS ESAL,
        1 AS ID
    FROM EMPLOYEE
    UNION ALL
    SELECT FIRST_VALUE(EMP.ESAL) OVER(ORDER BY EMP.ESAL DESC) AS ESAL,
        ID
    FROM EMPLOYEE EMP,
        (SELECT ESAL,
        ID+1 AS ID
        FROM CTE_NTH_SAL) CTE_NTH_SAL
        WHERE EMP.ESAL<CTE_NTH_SAL.ESAL
                AND CTE_NTH_SAL.ID<=5 )
    SELECT DISTINCT ESAL
FROM CTE_NTH_SAL
WHERE ID=5

for sample result set and more ways clickhere

Answer 7

Find Nth highest salary from a table. Here is a way to do this task using dense_rank() function.

select linkorder from u_links

select max(linkorder) from u_links

select max(linkorder) from u_links where linkorder < (select max(linkorder) from u_links)

select top 1 linkorder 
       from ( select distinct top 2 linkorder from u_links order by linkorder desc) tmp 
order by linkorder asc

DENSE_RANK : 1. DENSE_RANK computes the rank of a row in an ordered group of rows and returns the rank as a NUMBER. The ranks are consecutive integers beginning with 1. 2. This function accepts arguments as any numeric data type and returns NUMBER. 3. As an analytic function, DENSE_RANK computes the rank of each row returned from a query with respect to the other rows, based on the values of the value_exprs in the order_by_clause. 4. In the above query the rank is returned based on sal of the employee table. In case of tie, it assigns equal rank to all the rows.

WITH result AS ( 
     SELECT linkorder ,DENSE_RANK() OVER ( ORDER BY linkorder DESC ) AS  DanseRank 
FROM u_links ) 
SELECT TOP 1 linkorder FROM result WHERE DanseRank = 5

Answer 8

If you want optimize way means use TOP Keyword, So the nth max and min salaries query as follows but the queries look like a tricky as in reverse order by using aggregate function names:

N maximum salary:

SELECT MIN(EmpSalary)
FROM Salary
WHERE EmpSalary IN(SELECT TOP N EmpSalary FROM Salary ORDER BY EmpSalary DESC) 

for Ex: 3 maximum salary:

SELECT MIN(EmpSalary)
FROM Salary
WHERE EmpSalary IN(SELECT TOP 3 EmpSalary FROM Salary ORDER BY EmpSalary DESC) 

N minimum salary:

SELECT MAX(EmpSalary)
FROM Salary
WHERE EmpSalary IN(SELECT TOP N EmpSalary FROM Salary ORDER BY EmpSalary ASC)

for Ex: 3 minimum salary:

SELECT MAX(EmpSalary)
FROM Salary
WHERE EmpSalary IN(SELECT TOP 3 EmpSalary FROM Salary ORDER BY EmpSalary ASC)

Answer 9

Try this code :-

SELECT *
   FROM one one1
   WHERE ( n ) = ( SELECT COUNT( one2.salary )
                   FROM one one2
                   WHERE one2.salary >= one1.salary
                 )

Answer 10

To get third highest value from table

SELECT * FROM tableName ORDER BY columnName DESC LIMIT 2, 1

Answer 11

Subqueries always take more time:

use below query to get the any highest and lowest data:

Highest Data: select *from business order by id desc limit 3,1;

Lowest data: select *from business order by id asc limit 3,1;

Can use N in the place of 3 to get nth data.

Answer 12

NOTE: Please replace OFFSET 3 in Query with ANY Nth integer number

SELECT EmpName,EmpSalary
FROM SALARY
ORDER BY EmpSalary DESC
OFFSET 3 ROWS 
FETCH NEXT 1 ROWS ONLY

Description

FETCH NEXT 1 ROWS ONLY

return only 1 row

OFFSET 3 ROWS

exclude first 3 records Here you can you any integer number

Answer 13

SELECT EmpSalary 
FROM salary_table 
GROUP BY EmpSalary 
ORDER BY EmpSalary DESC LIMIT n-1, 1;

Answer 14

MySQL tested solution, assume N = 4:

select min(CustomerID) from (SELECT distinct CustomerID FROM Customers order by CustomerID desc LIMIT 4) as A;

Another example:

select min(country) from (SELECT distinct country FROM Customers order by country desc limit 3);

Answer 15

SELECT * FROM (select distinct Salary from Customers order by salary DESC) limit 4,1;

Limit 4,1 means leave first 4 rows and then select the next one.

Limit and rownumber depends on the platform you are using.

Try this,it will work.

Answer 16

select 
    Min(salary) 
from ( select salary from employees order by salary desc) t
where rownum<=3;

For 2nd highest salary,Change 3 to 2 in above query and for Nth highest salary to N where N = 1,2,3,4....

Answer 17

Try this one...

SELECT MAX(salary) FROM employee WHERE salary NOT IN (SELECT * FROM employee ORDERBY salary DESC LIMIT n-1)

Answer 18

Showing all 3rd highest salary:

select * from emp where sal=
(SELECT DISTINCT sal FROM emp ORDER BY sal DESC LIMIT 3,1) ;

Showing only 3rd highest salary:

SELECT DISTINCT sal FROM emp ORDER BY sal DESC LIMIT 3,1

Answer 19

Use ROW_NUMBER(if you want a single) or DENSE_RANK(for all related rows):

WITH CTE AS
(
    SELECT EmpID, EmpName, EmpSalary,
           RN = ROW_NUMBER() OVER (ORDER BY EmpSalary DESC)
    FROM dbo.Salary
)
SELECT EmpID, EmpName, EmpSalary
FROM CTE
WHERE RN = @NthRow

Answer 20

Optimized way: Instead of subquery just use limit.

select distinct salary from employee order by salary desc limit nth, 1;

See limit syntax here http://www.mysqltutorial.org/mysql-limit.aspx

Answer 21

--nth highest salary

select * 
from (select lstName, salary, row_number() over( order by salary desc) as rn 
      from employee) tmp
where rn = 2

--(nth -1) highest salary

select * 
from employee e1
where 1 = (select count(distinct salary)  
           from employee e2
           where e2.Salary > e1.Salary )

Answer 22

select min(salary) 
from (select salary 
      from employee 
      where rownum < n+1 
      order by salary desc);

Answer 23

Third or nth maximum salary from salary table without using subquery

select salary from salary
   ORDER   BY salary DESC
   OFFSET  N-1 ROWS
   FETCH NEXT 1 ROWS ONLY

For 3rd highest salary put 2 in place of N-1

Answer 24

SELECT MIN(COLUMN_NAME)
FROM   (
           SELECT DISTINCT TOP 3     COLUMN_NAME
           FROM   TABLE_NAME
           ORDER BY
                  COLUMN_NAME        DESC
       ) AS 'COLUMN_NAME'

Answer 25

By subquery:

SELECT salary from
(SELECT rownum ID, EmpSalary salary from
(SELECT DISTINCT EmpSalary from salary_table order by EmpSalary DESC)
where ID = nth)

Answer 26

set @n = $n

SELECT a.* FROM ( select a.* , @rn = @rn+1  from EMPLOYEE order by a.EmpSalary desc ) As a  where rn = @n

Answer 27

Method 1:

SELECT TOP 1 salary FROM (
SELECT TOP 3 salary 
 FROM employees 
  ORDER BY salary DESC) AS emp 
 ORDER BY salary ASC

Method 2:

  Select EmpName,salary from
  (
    select EmpName,salary ,Row_Number() over(order by salary desc) as rowid      
     from EmpTbl)
   as a where rowid=3

Answer 28

Refer following query for getting nth highest salary. By this way you get nth highest salary in MYSQL. If you want get nth lowest salary only you need to replace DESC by ASC in the query.

Answer 29

Select TOP 1 Salary as '3rd Highest Salary' from (SELECT DISTINCT TOP 3 Salary from Employee ORDER BY Salary DESC) a ORDER BY Salary ASC;

I am showing 3rd highest salary

Answer 30

Too simple if you use the sub query!

SELECT MIN(EmpSalary) from (
SELECT EmpSalary from Employee ORDER BY EmpSalary DESC LIMIT 3
);

You can here just change the nth value after the LIMIT constraint.

Here in this the Sub query Select EmpSalary from Employee Order by EmpSalary DESC Limit 3; would return the top 3 salaries of the Employees. Out of the result we will choose the Minimum salary using MIN command to get the 3rd TOP salary of the employee.