首先对于这道题的题目意思要理解清楚：不是说求除最大月之外近三个月的累计薪水；而是求每个月的近三个月的累计薪水，不足三个月也要计算。

方法一：

第一步：去掉每个员工，最近一个月的信息

-- with t as
-- (
-- select Employee.Id,Employee.Month,Employee.Salary
-- from Employee
-- inner join
-- (
-- select Id,max(Month) max_month
-- from Employee
-- group by Id
-- ) m
-- on Employee.Id=m.Id and
-- Employee.Month != m.max_month
-- )

 

第二步：使用左连接

-- select a.Id,a.Month,a.salary+ifnull(b.salary,0)+ifnull(c.salary,0) salary
-- from t a 
-- left join t b
-- on a.Id=b.Id and a.Month-b.Month=1
-- left join t c
-- on a.Id=c.Id and a.Month-c.Month=2
-- order by Id,Month desc;

方法二：

第一步：去掉每个员工，最近一个月的信息

select id,month,salary,rank() over(partition by id order by month desc) as rn 
from Employee；

第二步：

select id,month,sum(salary) over(partition by id order by month rows between 2 PRECEDING  and CURRENT ROW ) as salary 
from (
select id,month,salary,rank() over(partition by id order by month desc) as rn 
from Employee
) t where rn > 1 
order by id asc,month desc;

这里使用了row between 2 PRECEDING and CURRENT ROW

#题目意思理解错误
-- select Id,Month,
-- sum(Salary) over (partition by Id order by Month) salary
-- from
-- (
-- select Employee.Id,Employee.`Month`,Employee.Salary,a.mM-Employee.Month diffmonth
-- from Employee
-- left join
-- (
-- select Id,max(Month) mM
-- from Employee
-- group by Id
-- ) a
-- on Employee.Id=a.Id
-- ) b
-- where b.diffmonth in (1,2,3)
-- order by Id,Month desc;