Second Highest Salary - sql

Question

Write a SQL query to get the second highest salary from the Employee table.
| Id | Salary |
| 1 | 100 |
| 2 | 200 |
| 3 | 300 |
For example, given the above Employee table, the query should return 200 as the second highest salary. If there is no second highest salary, then the query should return null.
| SecondHighestSalary |
| 200 |
This is a question from Leetcode, for which I entered the following code:
SELECT CASE WHEN Salary = ''
THEN NULL
ELSE Salary
END AS SecondHighestSalary
FROM (SELECT TOP 2 Salary
,ROW_NUMBER() OVER (ORDER BY Salary DESC) AS Num
FROM Employee
ORDER BY Salary DESC) AS T
WHERE T.Num = 2
It says that the query does not return NULL if there's no value for second highest salary.
For eg. if the table is
| Id | Salary|
| 1 | 100 |
The query should return
|SecondHighestSalary|
| null |
and not
|SecondHighestSalary|
| |

Answer 1

Solution to the Leetcode 2nd highest salary problem is:
Select
Max(Salary) AS SecondHighestSalary
from Employee
where Salary < (
Select Max(Salary) from Employee
);

Answer 2

In case of ties you want the second highest distinct value. E.g. for values 100, 200, 300, 300, you want 200.
So get the highest value (MAX(salary) => 300) and then get the highest value less than that:
select max(salary) from mytable where salary < (select max(salary) from mytable);

Answer 3

Using window functions, utilizing NTH_VALUE gives a clean answer
SELECT (
SELECT NTH_VALUE(Salary, 2) OVER(ORDER BY Salary DESC) AS SecondHighestSalary
FROM Employee
GROUP BY Salary
LIMIT 1 OFFSET 1 ) AS SecondHighestSalary
;
Detailed Break-Down:
The outer SELECT Statement is required to get NULL value incase a value was not found (second rank is not present for example, or table only has one row)
NTH_VALUE(Salary, 2) is basically saying, look at each group (in our case it divides the table based on groups of Salary) and for each group add a column that lists the second highest value, for every row within the same group from that new column, we want to pick the second most paid (so only second row)
NTH_VALUE() OVER(ORDER BY) is in ASC order by default, make sure you explicit the DESC order
NTH_VALUE() merely gives the order to the rows within each group (here Salary) incase of two similar salaries in the same salary group it will give them separate ranks (ex 1 and 2) even if they have same value in the same group, For this use GROUP BY () statement
Because NTH_VALUE() merely gives the order to the columns, based on a group USE LIMIT 1 to get just one value (top value) and OFFSET 1 (to make that top value our targeted second most paid)

Answer 4

you should be able to do that with OFFSET 1/FETCH 1:
https://technet.microsoft.com/en-us/library/gg699618(v=sql.110).aspx

Answer 5

SELECT id, MAX(salary) AS salary
FROM employee
WHERE salary IN
(SELECT salary FROM employee MINUS SELECT MAX(salary)
FROM employee);
You can try above code to find 2nd maximum salary.
The above code uses MINUS operator.
For further reference use the below links
https://www.techonthenet.com/sql/minus.php
https://www.geeksforgeeks.org/sql-query-to-find-second-largest-salary/

Answer 6

You can use RANK() function to rank the values for Salary column, along with a CASE statement for returning NULL.
SELECT
CASE WHEN MAX(SalaryRank) = 1 THEN NULL ELSE Salary as SecondHighestSalary
FROM
(
SELECT *, RANK()OVER(ORDER BY Salary DESC) As SalaryRank
FROM Employee
) AS Tab
WHERE SalaryRank = 2
It would be better to use the DENSE_RANK() function so that ranks don't get skipped whenever there is a tie for a position.

Answer 7

I would use DENSE_RANK() & do LEFT JOIN with employee table :
SELECT t.Seq, e.*
FROM ( VALUES (2)
) t (Seq) LEFT JOIN
(SELECT e.*,
DENSE_RANK() OVER (ORDER BY Salary DESC) AS Num
FROM Employee e
) e
ON e.Num = t.Seq;

Answer 8

While you can use a CTE (from MSSQL 2005 or newer) or ROWNUMBER the easiest and more "portable" way is to just order by twice using a subquery.
select top 1 x.* from
(select top 2 t1.* from dbo.Employee t1 order by t1.Salary) as x
order by x.Salary desc
The requisite to show null when there's not a second bigger salary is a bit more tricky but also easy to do with a if.
if (select count(*) from dbo.Employee) > 1
begin
select top 1 x.* from
(select top 2 emp.* from dbo.Employee emp order by emp.Salary) as x
order by x.Salary desc
end
else begin
select null as Id, null as Salary
end
Obs:. OP don't said what to do when the second largest is a tie with the first but using this solution is a simple matter of using a DISTINCT in the IF subquery.

Answer 9

Here is the easy way to do this
SELECT MAX(Salary) FROM table WHERE Salary NOT IN (SELECT MAX(Salary) FROM table);

Answer 10

You can try this for getting n-th highest salary, where n = 1,2,3....(int)
SELECT TOP 1 salary FROM (
SELECT TOP n salary
FROM employees
ORDER BY salary DESC) AS emp
ORDER BY salary ASC
Hope this will help you. Below is one of the implementation.
create table #salary (salary int)
insert into #salary values (100), (200), (300)
SELECT TOP 1 salary FROM (
SELECT TOP 2 salary
FROM #salary
ORDER BY salary DESC) AS emp
ORDER BY salary ASC
drop table #salary
The output is here 200 as 300 is first highest, 200 is second highest and 100 is the third highest as shown below
salary
200
Here n is 2

Answer 11

Query:
CREATE TABLE a
([Id] int, [Salary] int)
;
INSERT INTO a
([Id], [Salary])
VALUES
(1, 100),
(2, 200),
(3, 300)
;
GO
SELECT Salary as SecondHighestSalary
FROM a
ORDER BY Salary
OFFSET 1 ROWS FETCH NEXT 1 ROWS ONLY
| SecondHighestSalary |
| ------------------: |
| 200 |

Answer 12

select case
when cnt>1 then SecondHighestSalary
else null end as SecondHighestSalary
from
(select top 1 Salary as SecondHighestSalary,
(select count(distinct Salary) from Employee) as cnt
from (
select distinct top 2 Salary
from Employee
order by Salary desc ) as sal
order by SecondHighestSalary asc) as b

Answer 13

Select salary from employees limit 1,1 ;
Very easy way to find second highest salary

Answer 14

select
case when max(salary) is null then null else max(salary) end SecondHighestSalary
from (
select salary , dense_rank() over (order by salary desc) as rn
from Employee
)r
where rn = 2
This code returns null when there is no second highest salary

Answer 15

You can use the union condition to handle the null case
SELECT Salary as "SecondHighestSalary" from Employee
WHERE Salary < (SELECT MAX(salary) FROM Employee )
UNION
(SELECT null)
ORDER BY 1 DESC
LIMIT 1;

Answer 16

You can use exists() together with If-else statements.
Query:
If exists(select distinct salary as SecondHighestSalary from employee
order by salary desc offset 1 row fetch first 1 row only)
select distinct salary as SecondHighestSalary from employee
order by salary desc offset 1 row fetch first 1 row only
else select null as SecondHighestSalary;

Answer 17

#Please check the below code#
SELECT TOP 1 secondhighestsalary
FROM (SELECT
CASE WHEN z.SalaryRank >1 THEN Salary
ELSE null END secondhighestsalary, SalaryRank
FROM
(SELECT salary, RANK() OVER(ORDER BY Salary DESC) As SalaryRank
FROM Employee GROUP BY salary )z
GROUP BY Salary, SalaryRank
)c
ORDER BY secondhighestsalary DESC
Explanation:
** SELECT salary, RANK() OVER(ORDER BY Salary DESC) As SalaryRank
FROM Employee GROUP BY salary**
above query is to provide rank to Salary column, group by clause will take care the duplicate values ..... next
**SELECT
CASE WHEN z.SalaryRank >1 THEN Salary
ELSE null END secondhighestsalary, SalaryRank
FROM
(SELECT salary, RANK() OVER(ORDER BY Salary DESC) As SalaryRank
FROM Employee GROUP BY salary )z **
next piece of code assign NULL against rank 1 and salary for greater than 1 rank.
So if you have only one row in the table and as per our question we need to display second highest salary if not display NULL, it will take care that situation.
At last we need to Order by DESC and take the Top 1 record.

Answer 18

SELECT max(Salary) as SecondHighestSalary from Employee where salary <>(SELECT max(salary) from Employee)