根据 SQL 中另一列中特定 ID 的条件计算持续时间

Question

我有一个名为 Audit 的表，如下所示：

ID	Record	AccessedDTTM
16	Sign In	2021-03-23 04:41:33.770
16	Session End - Sign Out	2021-03-23 04:42:33.770
17	Sign In	2021-03-01 00:03:04.070
17	Session End - Sign Out	2021-03-01 00:33:52.717
18	Sign In	2021-03-01 01:31:00.527
18	Schedule	2021-03-01 01:31:31.407
18	Charge	2021-03-01 01:43:27.427
18	Session End - Sign Out	2021-03-01 01:47:27.940

我需要计算 ID 为 16 和 18 的用户访问的持续时间。持续时间应基于他们的登录记录和会话结束 - 退出条目。

我试过这个查询： select DateDiff(second, (Select AccessedDTTM from Audit Where Record = 'Sign In'), (Select AccessedDTTM from Audit Where Record = 'Session End - Sign Out')) 作为持续时间 Foom 审计 ID 在哪里 (16, 18)

但我收到如下错误： 子查询返回超过 1 个值。当子查询跟随 =、!=、、>= 或子查询用作表达式时，这是不允许的。

有人可以帮助我进行正确的查询以实现此结果吗？提前致谢！

Answer 1

您可以使用条件聚合：

select id,
       datediff(second, 
                min(case when Record = 'Sign in' then AccessedDTTM end),
                max(case when Record = 'Session End - Sign Out' then AccessedDTTM end)
               ) as diff
from audit
where id in (16, 18)
group by id;

对数据做出一些合理的假设，您可以将其简化为：

select id,
       datediff(second, min(AccessedDTTM), max(AccessedDTTM)
               ) as diff
from audit
where id in (16, 18) and
      record in ('Sign in', 'Session End - Sign Out')
group by id;