使用dense_rank在MYSQL中找到第二高的薪水

Question

I am trying to do this Leetcode question where we try to find the 2nd highest salary from the EMPLOYEE TABLE.

INPUT:

+-------------+------+
| Column Name | Type |
+-------------+------+
| id          | int  |
| salary      | int  |
+-------------+------+

id is the primary key column for this table. Each row of this table contains information about the salary of an employee.

My answer is the following. But leetcode is not accepting this answer. What if I don't want to use the offset clause and use the dense function? Can it be done? Kindly help.

with temp1 as
(select id, salary, dense_rank() over (order by salary desc) as salary_order
from employee)
select coalesce(salary, null) as SecondHighestSalary
from temp1
where salary_order=2;

Question Link: https://leetcode.com/problems/second-highest-salary/

Answer 1

DENSE_RANK() is fine for this problem, but if you filter the cte and there is no 2nd highest salary, the query will not return null .
Also, coalesce(salary, null) would just return null if salary is already null , so it does not help.

With conditional aggregation you can get the 2nd highest salary and null if it does not exist:

WITH cte AS (SELECT salary, DENSE_RANK() OVER (ORDER BY salary DESC) AS salary_order FROM employee)
SELECT MAX(CASE WHEN salary_order = 2 THEN salary END) AS SecondHighestSalary
FROM cte;

See the demo .