如何在SQL Server中减去两个数据条目

问题描述

| 在我的数据库中，有一个表包含（ID，名称，时间，类型）

ID     Name    Time        Type
1      Osama   12:15 AM    IN
2      Osama   12:20 AM    OUT
3      Osama   14:15 AM    IN
4      Osama   14:20 AM    OUT

我需要构造一个查询以输出时差（OUT-IN）

Name,OUT-IN

例：

Osama,5
Osama,5

    

解决方法

这里的

TestData

CTE仅用于测试目的。另外，我注意到您的数据有误。最后我检查了“ 4”不是有效时间。它是

14:15

（通过24小时制）或

2:15 AM

（通过12小时制）。另外，此解决方案将需要SQL Server 2005或更高版本。

With TestData As
    (
    Select 1 As Id,\'Osama\' As Name,\'12:15\' As Time,\'IN\' As Type
    Union All Select 2,\'Osama\',\'12:20\',\'OUT\'
    Union All Select 3,\'14:15\',\'IN\'
    Union All Select 4,\'14:20\',\'OUT\'
    ),CheckInCheckOut As
    (
    Select Id,Name,Time,Type,Row_Number() Over ( Partition By Name,Type Order By Time ) As Num
    From TestData
    )
Select C1.Name,DateDiff( mi,CAST(C1.Time as datetime),Cast(C2.Time As datetime) ) As [OUT-IN]
From CheckInCheckOut As C1
    Join CheckInCheckOut As C2
        On C2.Name = C1.Name
            And C2.Type = \'OUT\'
            And C2.Num = C1.Num 
Where C1.Type = \'IN\'    

    ,如果可以确定您的时间是连续的，则可以执行Max（TimeRecorded）。这假设您的ID是连续的。您可以控制带有检查约束的任何一个。

declare @test table (
    Id int,Name varchar(50),TimeRecorded time,TypeOfTimeRecording varchar(3)
)

insert into @test values (1,CONVERT(time,\'12:15\'),\'IN\')
insert into @test values (2,\'12:20\'),\'OUT\')
insert into @test values (3,\'12:25\'),\'IN\')
insert into @test values (4,\'12:30\'),\'OUT\')

 select testOut.Name,testOut.TimeRecorded,testIn.TimeRecorded,DATEDIFF(minute,testOut.TimeRecorded) as [Out - In]
   from @test testOut
            inner join
        @test testIn    on testIn.Id = (select MAX(Id) from @test where Name = testOut.Name and Id < testOut.Id and TypeOfTimeRecording = \'IN\')
  where testOut.TypeOfTimeRecording = \'OUT\'