Sql 用户登录后立即发送消息的时间百分比是多少？

我以前从未问过这样的问题，也不知道这有多可能。假设我有下表：

user_id   date              event
22      2012-05-02 11:02:39 login
22      2012-05-02 11:02:53 send_message
22      2012-05-02 11:03:28 logout
22      2012-05-02 11:04:09 login
22      2012-05-02 11:03:16 send_message
22      2012-05-02 11:03:43 search_run

如何计算用户登录并在2分钟内发送消息的时间百分比？

对于给定用户：

SELECT round(count(*) FILTER (WHERE sent_in_time) * 100.0 / count(*), 2) AS pct_sent_in_time
FROM  (
   SELECT (min(date) FILTER (WHERE event = 'send_message')
         - min(date)) < interval '2 min' AS sent_in_time
   FROM  (
      SELECT date, event
           , count(*) FILTER (WHERE event = 'login')
                      OVER (ORDER BY date ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS grp
      FROM   tbl
      WHERE  user_id = 22  -- given user
      ) sub1
   GROUP  BY grp
   ) sub2;

对于所有用户：

SELECT user_id
     , round(count(*) FILTER (WHERE sent_in_time) * 100.0 / count(*), 2) AS pct_sent_in_time
FROM  (
   SELECT user_id
        , (min(date) FILTER (WHERE event = 'send_message')
         - min(date)) < interval '2 min' AS sent_in_time
   FROM  (
      SELECT user_id, date, event
           , count(*) FILTER (WHERE event = 'login')
                      OVER (PARTITION BY user_id ORDER BY date ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS grp
      FROM   tbl
      ) sub1
   GROUP  BY user_id, grp
   ) sub2
GROUP  BY user_id;

我扩展了测试用例以使其更具揭示性，因此使用了不同的百分比。见： 小提琴

在每次新登录后对数据进行分区，并检查“发送消息”是否在2分钟内发生。然后计算百分比和四舍五入

值得注意的是，这并没有被许多连续的登录所愚弄，在不到2分钟的时间里，我就用一条消息登录了

相关的：

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旁白：时间戳列的名称日期很容易引起误解。

在您的系统中是否可能出现欧文的小提琴中描述的场景？考虑了我能想到的几乎所有解决方案后，看起来与此非常相似。

SELECT user_id
     , round(count(*) FILTER (WHERE sent_in_time) * 100.0 / count(*), 2) AS pct_sent_in_time
FROM  (
   SELECT user_id
        , (min(date) FILTER (WHERE event = 'send_message')
         - min(date)) < interval '2 min' AS sent_in_time
   FROM  (
      SELECT user_id, date, event
           , count(*) FILTER (WHERE event = 'login')
                      OVER (PARTITION BY user_id ORDER BY date ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS grp
      FROM   tbl
      ) sub1
   GROUP  BY user_id, grp
   ) sub2
GROUP  BY user_id;

user_id | pct_sent_in_time ------: | ---------------: 22 | 33.33 23 | 100.00