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sql - 按月计算的累计值,填补缺失的月份

转载 作者:行者123 更新时间:2023-11-29 11:26:58 26 4
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我有这个数据表,我想知道是否可以创建一个查询,考虑到当前月份之前的所有月份,按月获取累计总和。

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,但最佳查询仍然很棘手。

基本查询快速获取运行总和:

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失的月份:

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+。这从表中的第一个月开始。
从给定月份开始:

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
, <b>COALESCE(</b>sum(c.mon_sum) OVER (ORDER BY mon)<b>, 0)</b> AS running_sum
FROM generate_series(<b>'2015-01-01'::date</b>, now(), interval '1 month') mon
LEFT JOIN cte c USING (mon)
ORDER BY mon;

db<> fiddle here
<子>旧sqlfiddle

将不同年份的月份分开。你没有要求,但你很可能会想要它。

请注意,“月份”在某种程度上取决于当前 session 的时区设置!详情:

相关:

关于sql - 按月计算的累计值,填补缺失的月份,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/32490946/

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