每个员工在进出房间时通常都有进出时间,有时会被搞砸,因为人们忘记了刷卡和尾随另一名员工。我想计算一下在房间里花的时间。
报告显示:
SELECT
u.userid
,u.[FirstName] + ' ' + u.[LastName] AS EmployeeName
,et.name AS [Description]
,MAX(e.LoggedTime) AS SwipeTime
,d.name AS door
FROM [Users] AS u
LEFT JOIN [Events] AS e ON e.RecordIndex1=u.UserID
LEFT JOIN [EventTypes] AS et on e.EventTypeID = et.EventTypeID
join [Doors] AS d ON e.RecordIndex2 = d.DoorID
WHERE LoggedTime > CONVERT(DATE, GETDATE()) and d.doorid in ( 32, 50, 42, 51, 33)
GROUP BY
u.userid,u.[FirstName] + ' ' + u.[LastName]
,et.name
,d.name
ORDER BY
u.[FirstName] + ' ' + u.[LastName]
,MAX(e.LoggedTime)我希望这份报告表明:
在时间列后面添加一列,并计算用户进出时间之间的时间差。
发布于 2019-07-19 14:39:15
通常,对于行之间的操作,延迟函数是goto解决方案。
LAG (scalar_expression [,offset] [,default])
OVER ( [ partition_by_clause ] order_by_clause ) 您可以使用它来引用前面(或后面使用铅)行。
https://stackoverflow.com/questions/57114301
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