Question

下表代表用户登录名（即LogAction_INT = 1是登录名，LogAction_INT = 0是注销）。汇总用户登录和注销（会话）之间的时间的最佳方法是什么。理想情况下，我需要每个用户总共花费时间。我能想到的一切都包括while循环，而且太复杂了。

ID  User_ID LogDate_DT  LogAction_INT
1940    18  2019-04-01 13:15:06.027 1
1941    18  2019-04-01 13:47:39.010 0
1942    18  2019-04-01 15:48:46.453 1
1943    18  2019-04-01 15:54:47.520 0
1944    68  2019-04-02 15:09:20.460 1
1945    68  2019-04-02 15:53:11.223 0
1946    86  2019-04-03 12:48:14.340 1
1947    86  2019-04-03 14:49:55.400 0
1948    80  2019-04-04 08:54:48.157 1
1949    86  2019-04-04 15:26:51.917 1
1950    86  2019-04-04 15:27:53.030 0
1951    86  2019-04-04 15:28:00.920 1
1952    86  2019-04-04 15:28:30.243 0
1953    86  2019-04-04 15:28:35.490 1
1954    86  2019-04-04 15:53:41.700 0
1955    68  2019-04-04 15:54:07.720 1
1956    80  2019-04-04 16:15:55.200 0

我希望有类似的东西：

User  TotalSessionTime
----  -----------------
18    04:45
68    10:02
80    06:12

Answer 1

如果它总是成对出现，则可以使用row_number（）生成运行中的no，然后将每2行分组为1

; with cte as
(
    select *, grp = (row_number() over (partition by User_ID order by ID) - 1) / 2
    from   your_table
)
cte2 as
(
    select  User_ID, elapsed = datediff(second, min(LogDate_DT), max(LogDate_DT))
    from    cte
    group by User_ID, grp
)
select User_ID, sum(elapsed)
from   cte2
group by User_ID

Answer 2

您可以枚举每种类型，然后使用条件聚合或联接：

select user_id, seqnum,
       datediff(second, min(LogDate_DT), max(LogDate_DT)) as diff_seconds
from (select t.*,
             row_number() over (partition by user_id, LogAction_INT order by id) as seqnum
      from t
     ) t
group by user_id, seqnum;

然后您可以按用户对它们进行求和：

select user_id, sum(diff_seconds)
from (select user_id, seqnum,
             datediff(second, min(LogDate_DT), max(LogDate_DT)) as diff_seconds
      from (select t.*,
                   row_number() over (partition by user_id, LogAction_INT order by id) as seqnum
            from t
           ) t
      group by user_id, seqnum;
     ) t
group by user_id;

这类问题的问题是，来龙去脉通常不那么整洁。这使这个问题变得更加棘手。

在受支持的SQL Server版本中，我将使用lag()进行此操作。

如何获得每个用户的标志之间的时间？

2 个答案: