我有下表叫做data_users
id | signed_up_at | product_id
-------+---------------------------------
20030 | 2017-09-15 12:51:45 | 2
20122 | 2017-09-15 12:51:45 | 2
21461 | 2017-09-15 12:51:45 | 2
20150 | 2017-09-13 10:10:10 | 2
19858 | 2017-09-10 23:00:54 | 2
20126 | 2017-09-10 23:00:28 | 2
20888 | 2017-09-10 23:00:28 | 2
20143 | 2017-09-10 23:00:28 | 2
21369 | 2017-09-10 23:00:02 | 2
我正在使用查询从该表中查找每天的累积总和:
SELECT DATE_TRUNC('day', signed_up_at::timestamptz) AS date,
SUM(COUNT(*)) OVER (ORDER BY DATE_TRUNC('day', signed_up_at::timestamptz)) AS sum
FROM "data_users"
WHERE product_id = 2
GROUP BY date ORDER BY date
这给了这样的东西
date | sum
------------------------+------
2017-09-15 00:00:00+02 | 1693
2017-09-13 00:00:00+02 | 1690
2017-09-10 00:00:00+02 | 1689
正如你所看到的那样,我在日期中存在空白,这些日期没有用户使用signed_up_at
。
而不是这个,我希望得到以下结果
date | sum
------------------------+------
2017-09-15 00:00:00+02 | 1693
2017-09-14 00:00:00+02 | 1690
2017-09-13 00:00:00+02 | 1690
2017-09-12 00:00:00+02 | 1689
2017-09-11 00:00:00+02 | 1689
2017-09-10 00:00:00+02 | 1689
因此,我希望每天都能获得总和,而不仅仅是那些拥有signed_up_at
用户的日子。
我试图用GENERATE_SERIES
实现这一点,但我没有得到预期的结果:
SELECT DATE_TRUNC('day', signed_up_at::timestamptz) AS date,
SUM(COUNT(*)) OVER (ORDER BY DATE_TRUNC('day', signed_up_at::timestamptz)) AS sum
FROM (SELECT GENERATE_SERIES(MIN(signed_up_at)::DATE, MAX(signed_up_at)::DATE, '1 DAY'::INTERVAL) AS date
FROM "data_users") AS d
LEFT OUTER JOIN "data_users" u ON u.signed_up_at::DATE = d.date::DATE
WHERE product_id = 2
GROUP BY signed_up_at ORDER BY date
我该如何修改我的查询以支持它? Turo的回答有助于指出我正确的方向,但我现在有WHERE
条款的问题。
我手头没有postgres,我看到伯爵和小组的问题,请试试
SELECT DATE_TRUNC('day', date::timestamptz) AS date,
SUM(COUNT(signed_up_at)) OVER (ORDER BY DATE_TRUNC('day',
date::timestamptz)) AS sum
FROM (SELECT GENERATE_SERIES(MIN(signed_up_at)::DATE, MAX(signed_up_at)::DATE,
'1 DAY'::INTERVAL) AS date
FROM "data_users") AS d
LEFT OUTER JOIN "data_users" u ON u.signed_up_at::DATE = d.date::DATE
GROUP BY date ORDER BY date