我有此数据表,我想知道是否有可能创建一个查询,考虑整个月的直到当前月份的,从而按月获取累计金额。
date_added | qty
------------------------------------
2015-08-04 22:28:24.633784-03 | 1
2015-05-20 20:22:29.458541-03 | 1
2015-04-08 14:16:09.844229-03 | 1
2015-04-07 23:10:42.325081-03 | 1
2015-07-06 18:50:30.164932-03 | 1
2015-08-22 15:01:54.03697-03 | 1
2015-08-06 18:25:07.57763-03 | 1
2015-04-07 23:12:20.850783-03 | 1
2015-07-23 17:45:29.456034-03 | 1
2015-04-28 20:12:48.110922-03 | 1
2015-04-28 13:26:04.770365-03 | 1
2015-05-19 13:30:08.186289-03 | 1
2015-08-06 18:26:46.448608-03 | 1
2015-08-27 16:43:06.561005-03 | 1
2015-08-07 12:15:29.242067-03 | 1
我需要这样的结果:
Jan|0
Feb|0
Mar|0
Apr|5
May|7
Jun|7
Jul|9
Aug|15
最佳答案
这与其他问题非常相似,但是最好的查询仍然很棘手。
基本查询以快速获取运行总和:
SELECT to_char(date_trunc('month', date_added), 'Mon YYYY') AS mon_text
, sum(sum(qty)) OVER (ORDER BY date_trunc('month', date_added)) AS running_sum
FROM tbl
GROUP BY date_trunc('month', date_added)
ORDER BY date_trunc('month', date_added);
棘手的部分是填写缺少的月份:
WITH cte AS (
SELECT date_trunc('month', date_added) AS mon, sum(qty) AS mon_sum
FROM tbl
GROUP BY 1
)
SELECT to_char(mon, 'Mon YYYY') AS mon_text
, sum(c.mon_sum) OVER (ORDER BY mon) AS running_sum
FROM (SELECT min(mon) AS min_mon FROM cte) init
, generate_series(init.min_mon, now(), interval '1 month') mon
LEFT JOIN cte c USING (mon)
ORDER BY mon;
implicit
CROSS JOIN LATERAL
需要Postgres 9.3+。从表中的第一个月开始。要从给定的月份开始:
WITH cte AS (
SELECT date_trunc('month', date_added) AS mon, sum(qty) AS mon_sum
FROM tbl
GROUP BY 1
)
SELECT to_char(mon, 'Mon YYYY') AS mon_text
, COALESCE(sum(c.mon_sum) OVER (ORDER BY mon), 0) AS running_sum
FROM generate_series('2015-01-01'::date, now(), interval '1 month') mon
LEFT JOIN cte c USING (mon)
ORDER BY mon;
SQL Fiddle.
将不同年份的月份间隔数月。您并没有要求,但是您很可能会想要它。
请注意,“月”在某种程度上取决于当前 session 的时区设置!细节:
有关的:
关于sql - 每月的累计值总和,填写缺少的月份,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/32490946/