sum over函数

sum over函数用于统计累计求和值

select name,mon,amount ,sum(amount) over (partition by name order by mon from(select name,mon,sum(amount) as amount from order group by name,mon )t 

sum(tota_amount)的求和是针对后面over()窗口的求和，over中partition by name order by mon 针对name这一组按照月份排序，

rows between unbounded preceding and current row 限定了行是按照在当前行不限定的往前处理，通俗就是处理当前以及之前的所有行的sum，即3月时sum(amount)求的时1、2、3月的和，2月时sum(amount)求的是1、2月的和。unbounded意思无限的 preceding在之前的，current row当前行。

换其他数据库一个 sum() over (partition by ... )就解决了..

mysql比较麻烦, 可以试下

select a.Daytime,sum(new_Role)

from (select distinct DATE_FORMAT(createtime,'%Y-%m-%d') as Daytime)

from `actors` WHERE ( (`createtime` >= '2019-02-01 00:00:00') AND (`createtime` <= '2019-02-11 24:00:00') )

) a

left join (select * from `actors`

WHERE ( (`createtime` >= '2019-02-01 00:00:00') AND (`createtime` <= '2019-02-11 24:00:00') )

)b

on a.Daytime>=b.createtime

group by a.Daytime

order by 1

累加当日、昨日、明日：

select a,b,sum(c) over(partition by a order by b rows between 1 preceding and 1 following） from t

累加当日和昨天：

select a,b,sum(c) over(partition by a order by b rows between 1 preceding and current row) from t

累加历史：分区内当天及之前所有

select a,b,sum(c) over(partition by a order by b) from t

或者：

select a,b,sum(c) over(partition by a order by b rows between unbounded preceding and current row) from t

累加分区内所有：当天和之前之后所有

select a,b,sum(c) over(partition by a order by b rows between unbounded preceding and unbounded following) from t

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