删除会话记录中的时间重叠

Question

我有一个 Sessions table 列（User_ID、Sessions_ID、LogOn、LogOut），用户可以同时打开多个会话，我的目标是计算每个用户在我的系统上花费的纯时间。我使用了以下查询：

SELECT
 T1.User_ID,
 SUM(T1.Duration) AS Duration
FROM (
 SELECT
  T2.User_ID,
  T2.Logon,
  (CASE WHEN T3.LogOn IS NULL OR T3.LogOn > T2.LogOut THEN T2.LogOut ELSE T3.LogOn END) AS LogOutEdited,
  DATEDIFF(MINUTE, T2.Logon, (CASE WHEN T3.LogOn IS NULL OR T3.LogOn > T2.LogOut THEN T2.LogOut ELSE T3.LogOn END)) AS Duration
 FROM (
  SELECT
   (DENSE_RANK() OVER (PARTITION BY User_ID ORDER BY LogOn)) AS Serial,
   User_ID, LogOn, LogOut
  FROM Sessions
 ) AS T2
 
 LEFT JOIN (
  SELECT
   (DENSE_RANK() OVER (PARTITION BY User_ID ORDER BY LogOn)) AS Serial,
   User_ID, LogOn, LogOut
  FROM Sessions
 ) AS T3
 ON T2.User_ID = T3.User_ID
  AND T2.Serial = T3.Serial - 1
) AS T1
GROUP BY T1.User_ID

此查询将一个会话的结束与下一个会话的开始进行比较，并调整第一个会话的结束以消除重叠时间。它确实给出了正确的结果（我认为 :) ）但是它的性能不被赞赏，我可以在这里应用更有效的逻辑吗？

编辑：

示例数据：

--------------------------------------------------------------------
| User_ID | Session_ID |        LogOn        |        LogOut       |
--------------------------------------------------------------------
|    1    |    100     | 2020-01-01 01:00:00 | 2020-01-01 01:30:00 |
--------------------------------------------------------------------
|    1    |    101     | 2020-01-01 01:15:00 | 2020-01-01 01:45:00 |
--------------------------------------------------------------------
|    1    |    102     | 2020-01-01 01:35:00 | 2020-01-01 01:40:00 |
--------------------------------------------------------------------
|    2    |    103     | 2020-01-01 03:13:00 | 2020-01-01 03:23:00 |
--------------------------------------------------------------------
|    1    |    104     | 2020-01-01 04:00:00 | 2020-01-01 04:15:00 |
--------------------------------------------------------------------

期望的结果：

----------------------
| User_ID | Duration |
----------------------
|    1    |    60    |
----------------------
|    2    |    10    |
----------------------

不想要的结果：

----------------------
| User_ID | Duration |
----------------------
|    1    |    80    |
----------------------
|    2    |    10    |
----------------------

Answer 1

这是一个间隙和孤岛问题，您要在其中尝试识别孤岛，并对每个用户的总持续时间求和。

这是一种使用 lag() 和 window sum() 来定义组的方法。以下查询为每组重叠会话提供一行：

select user_id, min(log_in) log_in, max(log_out) log_out
from (
    select 
        t.*,
        sum(case when log_in <= lag_log_out then 0 else 1 end) 
            over(partition by user_id order by log_in)  as grp
    from (
        select 
            t.*, 
            lag(log_out) over(partition by user_id order by log_in) as lag_log_out
        from mytable t
    ) t
) t
group by user_id, grp

您可以添加一级聚合来计算每个用户花费的总时间：

select user_id, sum(datediff(minute, login, log_out)) duration
from (
    select user_id, min(log_in) log_in, max(log_out) log_out
    from (
        select 
            t.*,
            sum(case when log_in <= lag_log_out then 0 else 1 end) 
                over(partition by user_id order by log_in)  as grp
        from (
            select 
                t.*, 
                lag(log_out) over(partition by user_id order by log_in) as lag_log_out
            from mytable t
        ) t
    ) t
    group by user_id, grp
) t
group by user_id

删除会话记录中的时间重叠

Remove Time Overlaps in Sessions Records

sql

sql-server

date

window-functions

gaps-and-islands