Question

我有一个包含所有日志文件的数据库供用户进入和离开办公室目前，我使用min和max函数计算每个用户每天的工作时间。

如果我需要计算用户留在办公室的确切时间，我可以从哪里开始？

目前我不知道如何开始。

并非每个人都有一个进出的记录。有时系统可能有2 IN或2 Out。如果发生这种情况，那么该日期的工作时间将为o。

    Date Occurred   Time Occurred   First Name  Last Name   Location

    9/1/2014        10:39:40        User A      User A      OfficeIn
    9/1/2014        12:34:36        User A      User A      OfficeOut
    9/1/2014        12:37:49        User A      User A      OfficeIn
    9/1/2014        12:39:51        User A      User A      OfficeOut
    9/1/2014        12:42:19        User A      User A      OfficeIn
    9/1/2014        14:09:32        User A      User A      OfficeIn
    9/1/2014        16:15:30        User A      User A      OfficeOut
    9/1/2014        16:17:40        User A      User A      OfficeIn
    9/1/2014        17:43:43        User A      User A      OfficeOut

Answer 1

因此，对于显示顺序中的记录，您希望查看上一条记录以计算时间跨度。 LAG是查看上一条记录的SQL函数。

对于上一条记录为OfficeIn的每个OfficeOut记录，我计算时差。然后我总结每个用户获得的所有时间部分。当你分开日期和时间（为什么???）时，我必须首先将它们组合起来。由于表格中没有人员ID，只有名字和姓氏，我们希望他们的组合是唯一的。

select 
  name,
  sum
  (
    case when location = 'OfficeOut'
    and lag(location) over(partition by name order by when) = 'OfficeIn' then
        datediff(minutes, lag(when) over(partition by name order by when)
    end
  ) as total_minutes
from
(
  select 
    concat(first_name, last_name) as name,
    location, 
    cast(date_occurred as datetime) + cast(time_occurred as datetime) as when
  from mytable
)
group by name
order by name;

现在进行错误检测。如果存在不一致，您希望总和为0。因此，我必须查找状态与上一条记录相同的记录。

select 
  name,
  case when max(case when location = lag(location) over(partition by name order by when) then 'ERROR' end) = 'ERROR' then 
    0
  else
    sum
    (
      case when location = 'OfficeOut'
      and lag(location) over(partition by name order by when) = 'OfficeIn' then
          datediff(minutes, lag(when) over(partition by name order by when)
      end
    )
  end as total_minutes
from
(
  select 
    concat(first_name, last_name) as name,
    location, 
    cast(date_occurred as datetime) + cast(time_occurred as datetime) as when
  from mytable
)
group by name
order by name;

Answer 2

检查以下链接，它可能对您有帮助 - 您将了解如何执行此操作 -

Working with Time Spans and Durations in SQL

SQL query for find out in/out time in a office

SQL计算时间在办公室

2 个答案: