我在构建聚合函数时遇到问题。我的问题是:
我有一张这样的桌子

 id action day         isSum difference
 1  ping   2012-01-01  1     500         (this is the sum of the differences from last year)
 2  ping   2012-01-01  0     -2
 3  ping   2012-01-02  0      1
 4  ping   2012-01-03  0     -4
 5  ping   2012-01-04  0     -2
 6  ping   2012-01-05  0      3
 7  ping   2012-01-06  0      2
 8  ping   2012-01-01  1      0         (this is the sum of the differences from last year, now for pong)
 9  pong   2012-01-01  0     -5
 10  pong   2012-01-02  0      2
 11  pong   2012-01-03  0     -2
 12  pong   2012-01-04  0     -8
 13  pong   2012-01-05  0      3
 14  pong   2012-01-06  0      4

我现在需要为每一天选择操作、日期和01-01以来的汇总差异,以便我的结果如下所示
action day        total
ping   2012-01-01 498
ping   2012-01-02 499
ping   2012-01-03 495
ping   2012-01-04 493
ping   2012-01-05 496
ping   2012-01-06 498
pong   2012-01-01 - 5
pong   2012-01-02 - 3
pong   2012-01-03 - 5
pong   2012-01-04 -13
pong   2012-01-05 -10
pong   2012-01-06 - 6

我该怎么做?
有很多数据集(~100万),所以查询需要非常便宜。我不知道如何使用sum根据action列获取每日记录的每日总和。

最佳答案

您需要使用SUBQUERY来获取total字段。

select action, day,
 (select sum(difference) from x
   where action = t2.action and day <= t2.day
   group by action) as total
from x t2 group by action ,day

SQLFIDDLE上查看

关于mysql - 建立基于SUM的每日记录,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/12454983/

10-15 03:20