累计天数 [英] Cumulative sum over days

问题描述

我有一个如下所示的MySQL表:

I have a MySQL table like the following:

date         count
2010-01-01   5
2010-01-02   6
2010-01-03   7

如何累积到下一天的每一天的总和?这样的结果是:

How can I accumulate the sum of each day to the next one? So the result is like:

date         acum per day
2010-01-01   5
2010-01-02   11
2010-01-03   18

我认为我需要某种for(每个日期)...但是没有任何线索.

I think i need some kind of for(each date)... but no clue.

最后一个查询我使用了Eric的回答. (谢谢).

Just the final query i used following answer from Eric. (thanks).

从

(SELECT count(*) operations, sum(amount), date(b.timestamp) dia
    FROM transactions b group by date(b.timestamp)) t1 

INNER JOIN 

(SELECT count(*) operations, sum(amount), date(b.timestamp) dia
    FROM transactions b group by date(b.timestamp)) t2 

ON t2.dia <= t1.dia GROUP BY t1.dia

推荐答案

好吧，我认为这行得通，但不确定性能如何:

Well, I think this would work, not sure how the performance would be though:

SELECT t1.date, sum(t2.count)
FROM mytable t1 INNER JOIN mytable t2 ON t2.date <= t1.date
GROUP BY t1.date

这篇关于累计天数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！