Employee 表保存了一年内的薪水信息。
请你编写 SQL 语句，来查询一个员工三个月内的累计薪水，但是不包括最近一个月的薪水。
结果请按 'Id' 升序，然后按 'Month' 降序显示。
示例：
输入：
| 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 |
输出：
| Id | Month | Salary |
|----|-------|--------|
| 1 | 3 | 90 |
| 1 | 2 | 50 |
| 1 | 1 | 20 |
| 2 | 1 | 20 |
| 3 | 3 | 100 |
| 3 | 2 | 40 |
解释：
员工 '1' 除去最近一个月（月份 '4'），有三个月的薪水记录：月份 '3' 薪水为 40，月份 '2' 薪水为 30，月份 '1' 薪水为 20。
所以近 3 个月的薪水累计分别为 (40 + 30 + 20) = 90，(30 + 20) = 50 和 20。
| Id | Month | Salary |
|----|-------|--------|
| 1 | 3 | 90 |
| 1 | 2 | 50 |
| 1 | 1 | 20 |
员工 '2' 除去最近的一个月（月份 '2'）的话，只有月份 '1' 这一个月的薪水记录。
| Id | Month | Salary |
|----|-------|--------|
| 2 | 1 | 20 |
员工 '3' 除去最近一个月（月份 '4'）后有两个月，分别为：月份 '4' 薪水为 60 和 月份 '2' 薪水为 40。所以各月的累计情况如下：
| Id | Month | Salary |
|----|-------|--------|
| 3 | 3 | 100 |
| 3 | 2 | 40 |
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/find-cumulative-salary-of-an-employee
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
leetcode 数据库题目全部题解
解法一
首先要明确“三个月内”是什么意思？
设当前月为a，则a，a-1，a-2为“三个月内”。
这种“前缀和”问题，通常采用自连接。
先找出每个月的“三个月内”的全部月份。
SELECT *
FROM employee AS E1 JOIN employee AS E2 ON (E1.Id = E2.Id AND E1.MONTH >= E2.MONTH and E2.MONTH >= E1.MONTH - 3)
其中：
E1.MONTH >= E2.MONTH and E2.MONTH >= E1.MONTH - 3
表示E1.MONTH的“三个月内”的全部月份。
再计算每个月三个月内的薪水累加和。
group by按id和month分组。计算E2.Salary的累加和。结果命名为A。
(
SELECT E1.Id,E1.MONTH,SUM(E2.Salary) AS `tsalary`
FROM employee AS E1 JOIN employee AS E2 ON (E1.Id = E2.Id AND E1.MONTH >= E2.MONTH and E2.MONTH >= E1.MONTH - 2)
GROUP BY E1.Id,E1.MONTH
) AS A
又因为不能包括最近一个月的薪水。
其实是要排除最高月份的值。
每个员工的最高月份。group by按id分组，最大月份值。
结果命名为B。
(
SELECT E.Id, MAX(E.MONTH) AS `max_month`
FROM employee AS E
GROUP BY E.Id
) AS B
结合表A和表B，选出每个员工最高月份之前的值。
结果请按 ‘Id’ 升序，按 ‘Month’ 降序显示 。
SELECT A.id,A.month,A.tsalary AS `Salary`
from
(
SELECT E1.Id,E1.MONTH,SUM(E2.Salary) AS `tsalary`
FROM employee AS E1 JOIN employee AS E2 ON (E1.Id = E2.Id AND E1.MONTH >= E2.MONTH and E2.MONTH >= E1.MONTH - 2)
GROUP BY E1.Id,E1.MONTH
) AS A
JOIN
(
SELECT E.Id, MAX(E.MONTH) AS `max_month`
FROM employee AS E
GROUP BY E.Id
) AS B
ON (A.Id=B.Id AND A.MONTH < B.max_month)
order by A.id,A.month desc
解法二
与解法一思路相同，既然只需要连续三个月的值。
干脆三个员工表自连接即可。
又由于1月份，2月份，向前不可能有三个月。
必然要用left join。
SELECT *
FROM employee AS E1
LEFT JOIN employee AS E2 ON (E1.Id = E2.Id AND E2.MONTH = E1.MONTH -1)
LEFT JOIN employee AS E3 ON (E3.Id = E2.Id AND E3.MONTH = E2.MONTH -1)
再讲三个月的薪水相加。注意到E2.Salary和E3.Salary为NULL时，薪水值为0.
结果命名为A。
FROM
(
SELECT E1.Id,E1.MONTH,(E1.Salary + if(E2.Salary IS NULL,0,E2.Salary) + if(E3.Salary IS NULL,0,E3.Salary)) AS `tsalary`
FROM employee AS E1
LEFT JOIN employee AS E2 ON (E1.Id = E2.Id AND E2.MONTH = E1.MONTH -1)
LEFT JOIN employee AS E3 ON (E3.Id = E2.Id AND E3.MONTH = E2.MONTH -1)
) AS A
求最大月份如解法一。
(
SELECT E.Id, MAX(E.MONTH) AS `max_month`
FROM employee AS E
GROUP BY E.Id
) AS B
之后同解法一，连接表A和表B，过滤掉最大月份值。
SELECT A.id,A.month,A.tsalary AS `Salary`
FROM
(
SELECT E1.Id,E1.MONTH,(E1.Salary + if(E2.Salary IS NULL,0,E2.Salary) + if(E3.Salary IS NULL,0,E3.Salary)) AS `tsalary`
FROM employee AS E1
LEFT JOIN employee AS E2 ON (E1.Id = E2.Id AND E2.MONTH = E1.MONTH -1)
LEFT JOIN employee AS E3 ON (E3.Id = E2.Id AND E3.MONTH = E2.MONTH -1)
) AS A
JOIN
(
SELECT E.Id, MAX(E.MONTH) AS `max_month`
FROM employee AS E
GROUP BY E.Id
) AS B
ON (A.Id=B.Id AND A.MONTH < B.max_month)
order by A.id,A.month desc
解法三
员工表按id升序，再按salary升序。
应用用户变量，可计算三个月内薪水累加和。
定义用户变量：@pre_id——上一行的id，@pre_salary1——前一个月的薪水值，@pre_salary2——前两个月的薪水值。
(SELECT @pre_id:= NULL,@pre_salary1:=0,@pre_salary2:=0) AS T
每个人，三个月内累计薪水的逻辑：
if (员工id != @pre_id){
三个月内累计薪水 = 员工id的薪水
}else{
三个月内累计薪水 = 员工id的薪水 + @pre_salary1 + @pre_salary2
}
SQL代码：
IF(E.Id != @pre_id,
E.Salary
,
E.Salary + @pre_salary1 + @pre_salary2
) AS `Salary`
用户变量更新逻辑：
if (员工id != @pre_id)
{
@pre_salary2 = 0
}
else{
@pre_salary2 = @pre_salary1
}
@pre_salary1 = 员工id的薪水
@pre_id = 员工id的id。
SQL代码：
IF(E.Id != @pre_id,
E.Salary
,
E.Salary + @pre_salary1 + @pre_salary2
) AS `Salary`,
if(E.Id != @pre_id,
@pre_salary2:=0
,
@pre_salary2:=@pre_salary1
) AS `pre2`,
(@pre_salary1:=E.Salary) AS `pre1`,
@pre_id:=E.Id
注意：上述逻辑成立的前提—— 员工表按id升序，再按salary升序。
将上述逻辑合起来，结果命名为表A。
(
SELECT
E.Id,
E.MONTH,
IF(E.Id != @pre_id,
E.Salary
,
E.Salary + @pre_salary1 + @pre_salary2
) AS `Salary`,
if(E.Id != @pre_id,
@pre_salary2:=0
,
@pre_salary2:=@pre_salary1
) AS `pre2`,
(@pre_salary1:=E.Salary) AS `pre1`,
@pre_id:=E.Id
FROM employee AS E,
(SELECT @pre_id:= NULL,@pre_salary1:=0,@pre_salary2:=0) AS T
ORDER BY E.Id,E.MONTH
) AS A
每个员工的最大月份类似解法一。
(
SELECT E.Id, MAX(E.MONTH) AS `max_month`
FROM employee AS E
GROUP BY E.Id
) AS B
之后同解法一，连接表A和表B，过滤掉最大月份值。
SELECT A.Id,A.MONTH,A.salary
FROM
(
SELECT
E.Id,
E.MONTH,
IF(E.Id != @pre_id,
E.Salary
,
E.Salary + @pre_salary1 + @pre_salary2
) AS `Salary`,
if(E.Id != @pre_id,
@pre_salary2:=0
,
@pre_salary2:=@pre_salary1
) AS `pre2`,
(@pre_salary1:=E.Salary) AS `pre1`,
@pre_id:=E.Id
FROM employee AS E,
(SELECT @pre_id:= NULL,@pre_salary1:=0,@pre_salary2:=0) AS T
ORDER BY E.Id,E.MONTH
) AS A
JOIN
(
SELECT E.Id, MAX(E.MONTH) AS `max_month`
FROM employee AS E
GROUP BY E.Id
) AS B
ON (A.Id=B.Id AND A.MONTH < B.max_month)
order by A.id,A.month desc
作者：jason-2
链接：https://leetcode-cn.com/problems/find-cumulative-salary-of-an-employee/solution/san-chong-jie-fa-by-jason-2/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。