问题
I have a MySQL table like the following:
date count
2010-01-01 5
2010-01-02 6
2010-01-03 7
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
I think i need some kind of for(each date)... but no clue.
Just the final query i used following answer from Eric. (thanks).
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
回答1:
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
回答2:
It's possible to solve this problem without join.
SET @cumulative_sum := 0;
SELECT date, @cumulative_sum := @cumulative_sum + count AS cumulative_sum
FROM table
ORDER BY date ASC
回答3:
Self join with a <= and sum the row would accomplish this.
来源：https://stackoverflow.com/questions/3095135/cumulative-sum-over-days