MySQL - Exclude Most Recent Month by Group - mysql

Question

Problem:
The Employee table holds the salary information in a year.
Write a SQL to get the cumulative sum of an employee's salary over a period of 3 months but exclude the most recent month.
The result should be displayed by 'Id' ascending, and then by 'Month' descending.
Employee table:
| Id | Month | Salary |
|----|-------|--------|
| 1 | 1 | 20 |
| 2 | 1 | 20 |
| 1 | 2 | 30 |
| 2 | 2 | 30 |
| 3 | 2 | 40 |
| 1 | 3 | 40 |
| 3 | 3 | 60 |
| 1 | 4 | 60 |
| 3 | 4 | 70 |
My Code:
SELECT t1.Id, t1.Month,
(SELECT SUM(Salary)
FROM Employee AS t2
WHERE t1.Id = t2.Id
AND t1.Month >= t2.Month) AS Salary
FROM Employee t1
WHERE Month <> (SELECT
MAX(Month)
FROM Employee
GROUP BY t1.Id)
ORDER BY Id, Month DESC;
My Output:
| Id | Month | Salary |
|----|-------|--------|
| 1 | 3 | 90 |
| 1 | 2 | 50 |
| 1 | 1 | 20 |
| 2 | 2 | 50 |
| 2 | 1 | 20 |
| 3 | 3 | 100 |
| 3 | 2 | 40 |
Expected:
| Id | Month | Salary |
|----|-------|--------|
| 1 | 3 | 90 |
| 1 | 2 | 50 |
| 1 | 1 | 20 |
| 2 | 1 | 20 |
| 3 | 3 | 100 |
| 3 | 2 | 40 |
I used MAX() and GROUP BY() functions to exclude the most recent month of each group, but it doesn't work for Id=2.
Is there any advice on how to get rid of the following row?
| 2 | 2 | 50 |
Thanks in advance.

Answer 1

To only get the cumulative sum for the last 3 months, excluding the most recent month per id, you can use
SELECT t1.Id, t1.Month, SUM(t2.Salary)
FROM Employee t1
JOIN Employee t2 ON t1.Id = t2.Id AND t1.Month - t2.Month <= 2 AND t1.Month - t2.Month >= 0
JOIN (SELECT id, MAX(month) as max_mth from Employee GROUP BY id) tmax on tmax.id=t1.id AND tmax.max_mth<>t1.month
GROUP BY t1.Id, t1.Month
ORDER BY t1.Id, t1.Month DESC;

Answer 2

Try this:
SELECT t1.id, t1.month,
(SELECT SUM(salary)
FROM employee t2
WHERE t1.id = t2.id
AND t1.month >= t2.month
AND t1.month - t2.month < 3) AS salary
FROM (
SELECT * FROM employee p
WHERE month <> (select MAX(month)
FROM employee c where c.id = p.id)) t1
ORDER BY id, month desc;
Output is:
+------+-------+--------+
| id | month | salary |
+------+-------+--------+
| 1 | 3 | 90 |
| 1 | 2 | 50 |
| 1 | 1 | 20 |
| 2 | 1 | 20 |
| 3 | 3 | 100 |
| 3 | 2 | 40 |
+------+-------+--------+
The problem you were having was that you were deleting only the last month present across all employees. What I believe you wanted was to delete the last month present for each employee even if that last month was several months back. This solution creates a derived table where the last month is missing for each employee and uses that in place of your t1 employee table.

Answer 3

I think this answer is closest to what you were trying to do in your original query:
SELECT t1.id, t1.month,
(SELECT SUM(salary)
FROM employee t2
WHERE t1.id = t2.id
AND t1.month >= t2.month
AND t1.month - t2.month < 3) AS salary
FROM employee t1
WHERE month <> (SELECT MAX(month)
FROM employee t3
WHERE t3.id = t1.id)
ORDER by id, month desc;
On second look you were actually pretty close. I believe the problem was that the "GROUP BY t1.Id" line doesn't actually group anything because t1.Id is constant for any given subquery as "t1" is defined in the outtermost select statement. Replace it with a where clause and limit the total to 3 months in the SUM() query, and you're there.

Answer 4

Try this query:
SELECT e.Id, e.Month, SUM( e2.Salary ) AS 'Salary'
FROM
Employee AS e
INNER JOIN Employee AS e2
ON e2.Id = e.Id
AND e2.Month <= e.Month
WHERE
e.Month <> ( SELECT MAX( [Month] ) FROM Employee WHERE Id = e.Id )
GROUP BY
e.Id, e.Month
ORDER BY
e.Id, e.Month DESC
Output is:
+----+-------+--------+
| Id | Month | Salary |
+----+-------+--------+
| 1 | 3 | 90 |
| 1 | 2 | 50 |
| 1 | 1 | 20 |
| 2 | 1 | 20 |
| 3 | 3 | 100 |
| 3 | 2 | 40 |
+----+-------+--------+