Sql 仅从Datetime列中减去时间

我有一个表中的两列，分别是

@ABC作为DateTime

和

@xyz作为DateTime

–数据类型

我只想减去时间，我试着这样做：

CREATE FUNCTION GetTimeDifference
(
@FirstDate datetime,
@SecondDate datetime
)
RETURNS varchar(10)
AS
BEGIN
DECLARE @Difference INT
DECLARE @FirstTimeInMin INT
DECLARE @SecondTimeInMin INT

SELECT @FirstTimeInMin = 
(DATEPART(hour,@FirstDate) * 60 + DATEPART(minute,@FirstDate))
SELECT @SecondTimeInMin =
(DATEPART(hour,@SecondDate) * 60 + DATEPART(minute,@SecondDate))

IF @FirstTimeInMin = 0
SET @FirstTimeInMin = 24 * 60

IF @SecondTimeInMin = 0
SET @SecondTimeInMin = 24 * 60

SET @Difference = @FirstTimeInMin - @SecondTimeInMin

IF(@Difference < 0)
SET @Difference = @Difference * -1

RETURN RIGHT('0' + CONVERT(varchar(10), @Difference / 60), 2)
 + ':' + 
RIGHT('0' + CONVERT(varchar(10), @Difference - (@Difference / 60) * 60 ), 2)

END
GO

SELECT dbo.GetTimeDifference('02/02/2012 6:10:00 PM','01/01/2001 12:00:00 AM')

对于

ABC=21/02/2012下午6:10:00

和

XYZ=01/01/2001上午12:00:00

->第一排

CONVERT(varchar(10), dbo.checkingtime.ABC – dbo.checkingtime.XYZ, 108)

我得到的结果是

18:10

，但我希望结果是

05:50

，时间仅为小时和分钟

---

可能吗

这是你想要的东西。这是内置函数，但我建议您构建自己的函数

DATEDIFF ( datepart , startdate , enddate )

请注意，您需要“屏蔽”同一天的开始和结束日期，以获得您想要的结果（这就是时间差）。结果将以分钟为单位，但您可以轻松地将其格式化为小时：分钟

---

干杯

您可以尝试以下功能：

CREATE FUNCTION GetTimeDifference
(
@FirstDate datetime,
@SecondDate datetime
)
RETURNS varchar(10)
AS
BEGIN
DECLARE @Difference INT
DECLARE @FirstTimeInMin INT
DECLARE @SecondTimeInMin INT

SELECT @FirstTimeInMin = 
(DATEPART(hour,@FirstDate) * 60 + DATEPART(minute,@FirstDate))
SELECT @SecondTimeInMin =
(DATEPART(hour,@SecondDate) * 60 + DATEPART(minute,@SecondDate))

IF @FirstTimeInMin = 0
SET @FirstTimeInMin = 24 * 60

IF @SecondTimeInMin = 0
SET @SecondTimeInMin = 24 * 60

SET @Difference = @FirstTimeInMin - @SecondTimeInMin

IF(@Difference < 0)
SET @Difference = @Difference * -1

RETURN RIGHT('0' + CONVERT(varchar(10), @Difference / 60), 2)
 + ':' + 
RIGHT('0' + CONVERT(varchar(10), @Difference - (@Difference / 60) * 60 ), 2)

END
GO

SELECT dbo.GetTimeDifference('02/02/2012 6:10:00 PM','01/01/2001 12:00:00 AM')

结果应该是05:50

这样就可以了

select right(date2-date1,7) as time from table1

结果:

TIME
5:50AM

---

我发现了一个很好的例子：

Select start_date, end_date, time_diff,
   EXTRACT(DAY FROM time_diff) days,
   EXTRACT(HOUR FROM time_diff) hours,
   EXTRACT(MINUTE FROM time_diff) minutes,
   EXTRACT(SECOND FROM time_diff) seconds
From
(
Select start_date, end_date, end_date - start_date time_diff
From
(
Select CAST(to_date('21/02/2012 06:10:00 am', 'dd/mm/yyyy  hh:mi:ss am') AS TIMESTAMP)  end_date
 , CAST(to_date('01/01/2012 12:00:00 am', 'dd/mm/yyyy  hh:mi:ss am') AS TIMESTAMP) start_date
From dual
))
/

---

从

6:10 pm

减去午夜应该得到

05:50

，这有什么意义？如果减法是另一种方式，这可能是有意义的……两个日期之间只需要几个小时的差吗？