Question

我有下表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

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

我正在使用查询从该表中查找每天的累计总和：

          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

这给出了类似的东西

signed_up_at

正如您所看到的那样，我的日期与没有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的用户的日期存在差距。

而不是这个，我希望得到以下结果

signed_up_at

所以我希望每天都能获得总和，而不仅仅是那些拥有GENERATE_SERIES用户的日子。

我正试图用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实现这一点，但我没有得到理想的结果：

WHERE

我应该如何修改我的查询以支持它？ Turo的回答有助于我指出正确的方向，但我现在遇到了import java.net.*; import java.io.*; class ScoketQuestion { public static void main(String[] args) { try { final String URL = ""; Socket s = new Socket(URL, 80); DataInputStream dIn = new DataInputStream(s.getInputStream()); PrintWriter wtr = new PrintWriter(s.getOutputStream()); wtr.println("GET / HTTP/1.1"); wtr.println("Host: " + URL); wtr.println(""); wtr.flush(); byte[] data = new byte[1024]; int pt = 0; String str; while (true) { pt = dIn.read(data); if (pt == -1) break; str = new String(data, 0, pt); System.out.println(str); } } catch (IOException e) { System.out.println(e); } } }条款的问题。

Answer 1

我手边没有任何违规行为，我看到计数和小组中的问题，请尝试

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

填补PostgreSQL查询结果中的空白

1 个答案: