Question

我有两列一个表@ABC as DateTime和@xyz as DateTime - 数据类型。

我想减去时间，我想这样做：

适用于ABC = 21/02/2012 6:10:00 PM和XYZ = 01/01/2001 12:00:00 AM - ＆gt;第一排。

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

我得到的结果为18:10，但我希望结果只在小时和分钟内显示为05:50。

有可能吗？？？

Answer 1

这是你想要的东西。这是内置的功能，但我会建议你建立自己的。

DATEDIFF ( datepart , startdate , enddate )

请注意，您需要在同一天“屏蔽”开始日期和结束日期以获得您想要的结果（这是时间差异）。结果将是分钟，但您可以轻松地将其格式化为小时：分钟。

欢呼声

Answer 2

您可以尝试这样的功能：

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

你可以像这样使用它：

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

结果应该是05:50

Answer 3

这样做，

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

结果：

TIME
5:50AM

SQL_LIVE_DEMO

Answer 4

这是我发现的好例子：

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
))
/

仅从日期时间列中减去时间

4 个答案: