How can I replace empty value to null with SQL? - mysql

Question

Let us say, I want to find the second highest salary. For that I use the following query.
select b.Salary as SecondHighestSalary from
(
select Salary, rank() over (order by Salary desc) as r
FROM Employee
) b
WHERE b.r = 2;
However, I want this to return null when the table only contains one entry, and therefore there is no 2nd highest salary. How can I do that?

Answer 1

Use your query with aggregation:
select max(Salary) as SecondHighestSalary
from (
select Salary, rank() over (order by Salary desc) as r
FROM Employee
) t
where r > 1;
The aggregate function max() will return null in case there is no 2nd highest salary.

Answer 2

One method would be:
select (select distinct salary
from employee
order by salary desc
limit 1 offset 1
);
The subquery returns NULL if the select distinct only returns one row.
With your structure, you can use conditional aggregation:
select max(case when seqnum = 2 then e.Salary end) as SecondHighestSalary
from (select Salary,
rank() over (order by Salary desc) as seqnum
from Employee e
) e;

Answer 3

Another easy hack with combination of union and limit.Below query will return null for no rows as well as single row.For more than one row it returns second largest salary.
select * from
(
select b.Salary as SecondHighestSalary from
(
select Salary, rank() over (order by Salary desc) as r
FROM Employee
) b
WHERE b.r = 2
union
select null) tab limit 1;
Refer DB Huddle link for full solution -https://dbfiddle.uk/?rdbms=mysql_8.0&fiddle=8b8b537160ca61e29da39a4662151217

Answer 4

Limit it to 1 result with rank of 1 or 2 and see if the rank is 2?
select if(c.r=2,c.Salary,NULL) SecondHighestSalary
from (
select b.Salary, b.r
from (
select Salary, rank() over (order by Salary desc) as r
FROM Employee
) b
where b.r <= 2
order by b.r
limit 1
) c
Even that will not return any results if there are no entries in Employee. If you want NULL in that case too, do:
select c.SecondHighestSalary
from (select 1) dummy
left join (
select b.Salary as SecondHighestSalary from
(
select Salary, rank() over (order by Salary desc) as r
FROM Employee
) b
WHERE b.r = 2
) c;

Answer 5

You can use the following:
WITH cteDesired_rank(desired_rank)
AS (SELECT 2)
select DISTINCT b.Salary as SecondHighestSalary
from cteDesired_rank rr
LEFT OUTER JOIN (select Salary, rank() over (order by Salary desc) as r
FROM employee) b
ON b.r = rr.desired_rank
WHERE rr.desired_rank = 2
Here we use a common table expression to specify the rank number that we want. We select first from the common table expression because we know that our desired rank number will be present. We then outer join to the rank subquery from employee where the desired rank number might be present, but is not required to be. We then filter for desired_rank = 2 - the desired_rank value will always be present, but the calculated rank value may not be. As such, if the subquery returns rows with the desired rank they will be displayed, but if the subquery does not return rows with the desired rank value the desired rank from the common table expression will still be returned, but the salary value will be NULL.
db<>fiddle here