use*_*259 4 mysql aggregate-functions
我有一个MySQL表,如下所示:
date count
2010-01-01 5
2010-01-02 6
2010-01-03 7
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我怎样才能将每一天的总和积累到下一个?结果如下:
date acum per day
2010-01-01 5
2010-01-02 11
2010-01-03 18
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我想我需要一些(每个日期)......但没有任何线索.
只是我在Eric的回答后使用的最终查询.(谢谢).
SELECT t1.dia,sum(t2.operacions),sum(t2.amount)FROM
(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
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嗯,我认为这会有效,但不确定性能如何:
SELECT t1.date, sum(t2.count)
FROM mytable t1 INNER JOIN mytable t2 ON t2.date <= t1.date
GROUP BY t1.date
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