How to get second-highest salary employees in a table

Question

It\'s a question I got this afternoon:

There a table contains ID, Name, and Salary of Employees, get names of the second-highest salary employees, in SQL Server

Answer 1

this is the simple query .. if u want the second minimum then just change the max to min and change the less than(<) sign to grater than(>).

    select max(column_name) from table_name where column_name<(select max(column_name) from table_name)

Answer 2

Using this SQL, Second highest salary will get with Employee Name

Select top 1 start at 2 salary from employee group by salary order by salary desc;

Answer 3

To get the names of the employees with the 2nd highest distinct salary amount you can use.

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

If Salary is indexed the following may well be more efficient though especially if there are many employees.

SELECT Name
FROM   Employees
WHERE  Salary = (SELECT MIN(Salary)
                 FROM   (SELECT DISTINCT TOP (2) Salary
                         FROM   Employees
                         ORDER  BY Salary DESC) T);

Test Script

CREATE TABLE Employees
  (
     Name   VARCHAR(50),
     Salary FLOAT
  )

INSERT INTO Employees
SELECT TOP 1000000 s1.name,
                   abs(checksum(newid()))
FROM   sysobjects s1,
       sysobjects s2

CREATE NONCLUSTERED INDEX ix
  ON Employees(Salary)

SELECT Name
FROM   Employees
WHERE  Salary = (SELECT MIN(Salary)
                 FROM   (SELECT DISTINCT TOP (2) Salary
                         FROM   Employees
                         ORDER  BY Salary DESC) T);

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

SELECT Name
FROM   Employees
WHERE  Salary = (SELECT DISTINCT TOP (1) Salary
                 FROM   Employees
                 WHERE  Salary NOT IN (SELECT DISTINCT TOP (1) Salary
                                       FROM   Employees
                                       ORDER  BY Salary DESC)
                 ORDER  BY Salary DESC)

SELECT Name
FROM   Employees
WHERE  Salary = (SELECT TOP 1 Salary
                 FROM   (SELECT TOP 2 Salary
                         FROM   Employees
                         ORDER  BY Salary DESC) sel
                 ORDER  BY Salary ASC)  

Answer 4

All of the following queries work for MySQL:

SELECT MAX(salary) FROM Employee WHERE Salary NOT IN (SELECT Max(Salary) FROM Employee);

SELECT MAX(Salary) From Employee WHERE Salary < (SELECT Max(Salary) FROM Employee);

SELECT Salary FROM Employee ORDER BY Salary DESC LIMIT 1 OFFSET 1;

SELECT Salary FROM (SELECT Salary FROM Employee ORDER BY Salary DESC LIMIT 2) AS Emp ORDER BY Salary LIMIT 1;

Answer 5

select max(age) from yd where age<(select max(age) from HK) ; /// True two table Highest 

SELECT * FROM HK E1 WHERE 1 =(SELECT COUNT(DISTINCT age) FROM HK E2 WHERE E1.age < E2.age); ///Second Hightest age RT single table 

select age from hk e1 where (3-1) = (select count(distinct (e2.age)) from yd e2 where e2.age>e1.age);//// same True Second Hight age RT two table

select max(age) from YD where age not in (select max(age) from YD);  //second hight age in single table 

Answer 6

This might help you

SELECT 
      MIN(SALARY) 
FROM 
      EMP 
WHERE 
      SALARY in (SELECT 
                      DISTINCT TOP 2 SALARY 
                 FROM 
                      EMP 
                 ORDER BY 
                      SALARY DESC
                )

We can find any nth highest salary by putting n (where n > 0) in place of 2

Example for 5^th highest salary we put n = 5