Question

我为用户提供了以下操作序列，并为每个操作设置了DateTime时间戳。下面是一个示例表

Actions:
+------------+-------------+----------------------+
| session_id | action_name | time                 | 
+------------+-------------+----------------------+
| 123abcd    | ADD         | 2014-08-27 13:41:02  |
+------------+-------------+----------------------+
| 123abcd    | LIKE        | 2014-08-27 13:43:02  |
+------------+-------------+----------------------+
| 123abcd    | DELETE      | 2014-08-27 13:45:02  |
+------------+-------------+----------------------+
| 123abcd    | EMAIL       | 2014-08-27 13:48:02  |
+------------+-------------+----------------------+
| 123abcd    | LIKE        | 2014-08-27 13:52:02  |
+------------+-------------+----------------------+
| 546erqr    | ADD         | 2014-08-27 14:02:02  |
+------------+-------------+----------------------+
| 546erqr    | EMAIL       | 2014-08-27 14:12:02  |
+------------+-------------+----------------------+
| 546erqr    | LIKE        | 2014-08-27 14:14:02  |
+------------+-------------+----------------------+

正如您所看到的，user可以通过session执行某些操作。

我对计算用户为每个会话发送电子邮件所花费的时间感兴趣。

基本上，我想要TIMESTAMPDIFF(SECOND, min(time), max(time))/60直到action = EMAIL。我尝试在SQL语句中使用where子句执行它，但它没有工作。

对于输出，我想要类似的东西：

+------------+-----------------------+
| session_id | time_till_email_mins  | 
+------------+-----------------------+
| 123abcd    | 7                     |
+------------+-----------------------+
| 546erqr    | 10                    |
+------------+-----------------------+

一般来说，如果要做这样的计算（无论是时间差计算还是计算某些事情）直到某个动作，那么如何进行计算。

希望这是有道理的。如果您有任何问题，请告诉我

Answer 1

您可以使用条件聚合执行此操作：

select session_id,
       timestampdiff(second, min(case when action_name = 'EMAIL' then action_time end),
                     min(action_time)) / 60
from actions a
group by session_id;

Answer 2

我想你可能想要这个：

select   session_id,
         min(time) as first_action_time,
         min(case when action_name = 'EMAIL' then time end) as first_email_time,
         timestampdiff(second,min(time),min(case when action_name = 'EMAIL' then time end)) as diff_in_secs
from     actions
group by session_id

这是会话的第一个操作与操作名称为＆＃39; EMAIL＆＃39;的会话的第一个操作之间的秒数差异。

Answer 3

使用子查询获取两个值。时间不是一个好的专栏名称。

select session_ID , **(datediff function using first_time and end_time)**

from
(select session_id , min(time) as first_time 
from actions 
group by session)a

inner join

(select session_id, time as end_time
from actions
where action_name = 'email'
group by session_ID) b

on a.session_id = b.session_id

不幸的是我不熟悉MySQL中的datediff / time diff函数（我把占位符放在选择行中），但你应该可以从这里开始工作。

MySQL获得持续时间到选定的操作

3 个答案: