Question

我有一个带有4个（相关列）的数据集：唯一ID，user_id，time_stamp，event。唯一ID是主键，user_id可以重复，time_stamp（datetime）在事件发生时采取，事件是a）推送通知（推送）或b）用户打开应用程序（打开）。

它看起来像这样：

id    user_id    time_stamp    event  count it?
 1        1          10         open
 2        1          23         push   -good
 3        1          28         open 
 4        1          38         push   -bad
 5        1          65         open   
 6        1          85         push   -good
 7        1          89         open   
 8        1          28         push   -bad
 9        2          38         push   -good
10        2          45         open
11        2          46         open

我想知道我的推送通知是否有用。为此，我需要查看用户是否在推送通知的20分钟内打开应用程序。我认为这是一次成功的推动＆＃34;而所有其他推动都不会成功。到目前为止，我已经有了内心加入表格的想法，但我遇到了重复行的问题。例如，我们会使用ID 4获得误报，因为只应计算ID 3。

  SELECT * FROM
  (SELECT * FROM table WHERE row = 'open') a
  INNER JOIN (SELECT * FROM table WHERE row = 'push') b 
  ON a.user_id = b.user_id) WHERE a.time_stamp - b.timestamp < 20;

Answer 1

因为你有同一个user_id的多条记录，我认为你想要采用最新的'open'time_stamp并将其与每个用户的最新'push'进行比较？

如果是这样，我认为以下内容符合您的要求（需要整理，但应该做到这一点）：

SELECT et4.User_id, ts1, et3.User_id, ts2
FROM 
(SELECT et1.user_id, max(et1.time_stamp) as ts1 from eventtable as et1 
where et1.event = 'push' 
group by et1.user_id
) as et4
INNER JOIN 
(SELECT et2.user_id, max(et2.time_stamp) as ts2 from eventtable as et2 
where event = 'open' group by et2.user_id) as et3 
ON et3.user_id = et4.user_id
WHERE ts2 -ts1 < 20

基本上，为每个用户选择最新推送，并将其连接到该用户的最新打开，然后计算时间戳的差异。

我希望这会有所帮助。

Answer 2

您可以尝试这样的事情：

SELECT t1.id, t1.user_id, t1.time_stamp, t1.event,
       t2.id, t2.time_stamp, t2.event     
FROM mytable AS t1
INNER JOIN mytable AS t2 
   ON t1.user_id = t2.user_id AND t1.event = 'push' AND  t2.event = 'open' AND 
      t2.time_stamp > t1.time_stamp AND t2.time_stamp - t1.time_stamp < 20
LEFT JOIN mytable AS t3 
   ON t3.user_id = t2.user_id AND t3.event = 'open' AND 
      t3.time_stamp > t1.time_stamp AND t3.time_stamp < t2.time_stamp
WHERE t3.id IS NULL

<强>输出：

id, user_id, time_stamp, event, id, time_stamp, event
=====================================================
2,  1,       23,         push,  3,   28,        open
8,  2,       28,         push,  10,  45,        open
9,  2,       38,         push,  10,  45,        open

注意：如果您需要使用LEFT JOIN过滤掉记录，则需要额外的id = 8。

计算时间之间的差异

2 个答案: