大数据第61天—MySQL之员工累计薪水-杨大伟

需求：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。

展示效果：

Id	Month	Salary
1	3	90
1	2	50
1	1	20
2	1	20
3	3	100
3	2	40

Create table If Not Exists 19_Employee (Id int, Month int, Salary int);
Truncate table 19_Employee;
insert into 19_Employee (Id, Month, Salary) values (1, 1, 20);
insert into 19_Employee (Id, Month, Salary) values (2, 1, 20);
insert into 19_Employee (Id, Month, Salary) values (1, 2, 30);
insert into 19_Employee (Id, Month, Salary) values (2, 2, 30);
insert into 19_Employee (Id, Month, Salary) values (3, 2, 40);
insert into 19_Employee (Id, Month, Salary) values (1, 3, 40);
insert into 19_Employee (Id, Month, Salary) values (3, 3, 60);
insert into 19_Employee (Id, Month, Salary) values (1, 4, 60);
insert into 19_Employee (Id, Month, Salary) values (3, 4, 70);

说明：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。

最终SQL:

SELECT
    E1.id,
    E1.month,
    (IFNULL(E1.salary, 0) + IFNULL(E2.salary, 0) + IFNULL(E3.salary, 0)) AS Salary
FROM
    (SELECT
        id, MAX(month) AS month
    FROM
        19_Employee
    GROUP BY 
        id
    HAVING 
        COUNT(*) > 1) AS maxmonth
    LEFT JOIN
        19_Employee E1 
    ON 
        (maxmonth.id = E1.id AND maxmonth.month > E1.month)
    LEFT JOIN
        19_Employee E2 
    ON 
        (E2.id = E1.id AND E2.month = E1.month - 1)
    LEFT JOIN 
        19_Employee E3 
    ON
        (E3.id = E1.id AND E3.month = E1.month - 2)
ORDER BY 
    id ASC , month DESC;