Question

我正在构建一个WCF应用程序，用于计算进出时间之间的总时间，从数据库中获取数据我使用GROUP BY子句按日期对数据进行分组，但我想要我的开始的日子＆amp;在早上6点结束，所以如果有人在早上3点离开，它将仅在当天加入。我正在使用以下命令查询

SELECT MIN([Swipedatetime]) AS [Entry]
     , MAX([Swipedatetime]) AS [Exit]
     , [UserID]
  FROM [Database_Name].[dbo].[Table_Name]
 where UserID = '100'
 GROUP 
    BY UserID
     , CAST (Swipedatetime as DATE)
 ORDER 
    BY MIN([Swipedatetime])

此外，如果有任何方法可以在存储过程中计算两次之间的差异，那么请提及它，它将是非常有用的。

Answer 1

要在早上6点到下午6点之间获取记录，您可以使用：

where datepart(hour,[Swipedatetime]) > 6 
and datepart(hour,[Swipedatetime]) <=18

对于差异你可以使用它：

select DATEDIFF(minute, MIN([Swipedatetime]), MAX([Swipedatetime]))

完全查询：

    declare @StartDate datetime = dateadd(HH, 6, convert(datetime, convert(date, getdate())))
    declare @EndDate datetime = dateadd(day,1,@Startdate)

SELECT MIN([Swipedatetime]) AS [Entry]
     , MAX([Swipedatetime]) AS [Exit]
     , DATEDIFF(minute, MIN([Swipedatetime]), MAX([Swipedatetime])) AS[Diff]
     , [UserID]
  FROM [Database_Name].[dbo].[Table_Name]
 where UserID = '100'
and [Swipedatetime] >= @Startdate
and [Swipedatetime] < @EndDate
 GROUP 
    BY UserID
     , CAST (Swipedatetime as DATE)
 ORDER 
    BY MIN([Swipedatetime])

Answer 2

如何从Swipedatetime中扣除6个小时并按新值分组：

GROUP BY (Swipedatetime - INTERVAL '6 hours')

（这是postgresql，对于sql-server我认为你需要函数dateadd（小时，-6，Swipedatetime），或者沿着这条线的东西）

Answer 3

您的解决方案只需要简单的DATEADD功能：

SELECT MIN([Swipedatetime]) AS [Entry]
     , MAX([Swipedatetime]) AS [Exit]
     , [UserID]
  FROM [dbo].[Table_Name]
 WHERE UserID = '100'
 GROUP 
    BY UserID
     , CAST (DATEADD(HOUR,6,Swipedatetime) AS DATE)
 ORDER 
    BY MIN([Swipedatetime])

分组根据自定义时间段查询日期

3 个答案: