34

我在计算两个日期之间的时差时遇到了一些困难。

我想要的是，我有两个约会可以说

@StartDate = '10/01/2012 08:40:18.000'
@EndDate='10/04/2012 09:52:48.000'

所以两个日期之间的差的形式hh:mm:ss是72:42:30。

如何在 T-SQL 查询中获得此结果？

4

14 回答 14

59 
declare @StartDate datetime, @EndDate datetime

select @StartDate = '10/01/2012 08:40:18.000',@EndDate='10/04/2012 09:52:48.000'

select convert(varchar(5),DateDiff(s, @startDate, @EndDate)/3600)+':'+convert(varchar(5),DateDiff(s, @startDate, @EndDate)%3600/60)+':'+convert(varchar(5),(DateDiff(s, @startDate, @EndDate)%60)) as [hh:mm:ss]

此查询将对您有所帮助。

于 2012-11-27T06:26:46.727 回答
36

最短的代码是：

Select CAST((@EndDateTime-@StartDateTime) as time(0)) '[hh:mm:ss]'
于 2014-09-03T16:30:17.600 回答
9

虽然可能不是最有效的，但这会起作用：

declare @StartDate datetime, @EndDate datetime

select @StartDate = '10/01/2012 08:40:18.000',@EndDate='10/04/2012 09:52:48.000'

select convert(varchar(5),DateDiff(s, @startDate, @EndDate)/3600)+':'+convert(varchar(5),DateDiff(s, @startDate, @EndDate)%3600/60)+':'+convert(varchar(5),(DateDiff(s, @startDate, @EndDate)%60))

如果您可以运行两个选择，那么这会更好，因为您只执行一次 datediff：

declare @StartDate datetime, @EndDate datetime

select @StartDate = '10/01/2012 08:40:18.000',@EndDate='10/04/2012 09:52:48.000'
declare @Sec BIGINT

select @Sec = DateDiff(s, @startDate, @EndDate)

select convert(varchar(5),@sec/3600)+':'+convert(varchar(5),@sec%3600/60)+':'+convert(varchar(5),(@sec%60))
于 2012-11-27T06:00:07.480 回答
8

我喜欢把它变成一个函数的想法，这样它就可以重复使用，并且你的查询变得更容易阅读：

--get the difference between two datetimes in the format: 'h:m:s'
CREATE FUNCTION getDateDiff(@startDate DATETIME, @endDate DATETIME)
RETURNS VARCHAR(10)
AS BEGIN
    DECLARE @seconds INT = DATEDIFF(s, @startDate, @endDate)
    DECLARE @difference VARCHAR(10) =
    CONVERT(VARCHAR(4), @seconds / 3600) + ':' +
    CONVERT(VARCHAR(2), @seconds % 3600 / 60) + ':' +
    CONVERT(VARCHAR(2), @seconds % 60)
    RETURN @difference
END

用法：

DECLARE @StartDate DATETIME = '10/01/2012 08:40:18.000'
DECLARE @endDate DATETIME = '10/04/2012 09:52:48.000'

SELECT dbo.getDateDiff(@startDate, @endDate) AS DateDifference

结果：

    DateDifference
1   73:12:30

如果添加填充也更容易阅读结果，因此格式始终为hh:mm:ss. 例如，以下是在 SQL Server 2012 或更高版本中执行此操作的方法：

--get the difference between two datetimes in the format: 'hh:mm:ss'
CREATE FUNCTION getDateDiff(@startDate DATETIME, @endDate DATETIME)
RETURNS VARCHAR(10)
AS BEGIN
    DECLARE @seconds INT = DATEDIFF(s, @startDate, @endDate)
    DECLARE @difference VARCHAR(10) =
    FORMAT(@seconds / 3600, '00') + ':' +
    FORMAT(@seconds % 3600 / 60, '00') + ':' +
    FORMAT(@seconds % 60, '00')
    RETURN @difference
END

请注意，如果小时数超过 2 位，这将不会缩短小时数。因此 1 小时将显示为01:00:00100 小时将显示为100:00:00

于 2017-01-09T16:49:44.927 回答
4

如果您不反对隐式类型转换，我将提供另一种解决方案。更好的格式是否更具可读性？你来做法官。

DECLARE  @StartDate datetime = '10/01/2012 08:40:18.000'
        ,@EndDate   datetime = '10/04/2012 09:52:48.000'

SELECT
    STR(ss/3600, 5) + ':' + RIGHT('0' + LTRIM(ss%3600/60), 2) + ':' + RIGHT('0' + LTRIM(ss%60), 2) AS [hh:mm:ss]
FROM (VALUES(DATEDIFF(s, @StartDate, @EndDate))) seconds (ss)
于 2014-03-14T16:33:14.593 回答
3 
DECLARE @dt1 datetime='2012/06/13 08:11:12', @dt2 datetime='2012/06/12 02:11:12'

SELECT CAST((@dt2-@dt1) as time(0))
于 2014-05-08T14:19:49.797 回答
3

如果您需要两个日期之间的零填充差异：

SELECT convert(varchar(2),FORMAT(DATEDIFF(s, @startDate, @endDate)/3600,'0#'))+':'    
   +convert(varchar(2),FORMAT(DATEDIFF(s, @startDate, @endDate)%3600/60,'0#'))+':'
   +convert(varchar(2),FORMAT(DATEDIFF(s, @startDate, @endDate)%60,'0#')) AS Duration
于 2019-07-11T12:07:40.167 回答
1

试试这个：

declare @StartDate datetime, @EndDate datetime

select @StartDate = '2016-05-04 10:23:41.083', @EndDate='2016-05-04 10:25:26.053'

select CAST(DateDiff(MI, @startDate, @EndDate)/60 AS varchar)+':'+Cast(DateDiff(MI, @startDate, @EndDate)%60 AS varchar)+':'+cast(DateDiff(s, @startDate, @EndDate)%60 AS varchar) as [hh:mm:ss]
于 2016-05-09T06:29:03.880 回答
0

看看这些。我没有使用更多的括号来保持可读性，所以请记住，乘法是在加法或减法之前完成的。

以下均返回：

hr  mins  sec   timediff
73   12    30   73:12:30

这是为了不使用子查询而编写的，并且是最易读易懂的： 

declare @StartDate datetime,
@EndDate datetime

set @StartDate = '10/01/2012 08:40:18.000'
set @EndDate =   '10/04/2012 09:52:48.000' 

select datediff(hour, @StartDate, @EndDate) hr,
   datediff(minute, @StartDate, @EndDate) 
   - datediff(hour, @StartDate, @EndDate) * 60 mins,
   datediff(second, @StartDate, @EndDate) 
   - (datediff(minute, @StartDate, @EndDate) * 60) sec,
   cast(datediff(hour, @StartDate, @EndDate) as varchar)+':'+ 
   cast(datediff(minute, @StartDate, @EndDate) 
   - datediff(hour, @StartDate, @EndDate) * 60 as varchar)+':'+ 
   cast(datediff(second, @StartDate, @EndDate) 
   - (datediff(minute, @StartDate, @EndDate) * 60) as varchar) timediff

如果您有大量数据，此版本的性能会更好。它需要一个子查询。

declare @StartDate datetime,
@EndDate datetime

set @StartDate = '10/01/2012 08:40:18.000'
set @EndDate =   '10/04/2012 09:52:48.000' 

select s.seconds / 3600  hrs,
s.seconds / 60 - (seconds / 3600 ) * 60 mins,
s.seconds - (s.seconds / 60) * 60   seconds,
cast(s.seconds / 3600 as varchar) + ':' +
cast((s.seconds / 60 - (seconds / 3600 ) * 60) as varchar) + ':' +
cast((s.seconds - (s.seconds / 60) * 60) as varchar) timediff
from (select datediff(second, @StartDate, @EndDate) as seconds) s
于 2012-11-27T07:05:01.447 回答
0

我今天遇到了这篇文章，因为我试图收集位于不同表中的字段之间的时差，这些表在一​​个关键字段上连接在一起。这是这种努力的工作代码。（在 sql 2010 中测试）请记住，我的原始查询在一个公共键域上共同加入了 6 个表，在下面的代码中，我删除了其他表，以免给读者造成任何混淆。

查询的目的是计算变量 CreatedUTC 和 BackupUTC 之间的差异，其中差异以天数表示，该字段称为“DaysActive”。

declare @CreatedUTC datetime
declare @BackupUtc datetime


SELECT TOP 500

table02.Column_CreatedUTC AS DeviceCreated,
CAST(DATEDIFF(day, table02.Column_CreatedUTC, table03.Column_EndDateUTC) AS nvarchar(5))+ ' Days' As DaysActive,
table03.Column_EndDateUTC AS LastCompleteBackup

FROM

Operations.table01 AS table01

LEFT OUTER JOIN

    dbo.table02 AS table02
ON
    table02.Column_KeyField = table01.Column_KeyField

LEFT OUTER JOIN 

    dbo.table03 AS table03
ON
    table01.Column_KeyField = table03.Column_KeyField

Where table03.Column_EndDateUTC > dateadd(hour, -24, getutcdate()) --Gathers records with an end date in the last 24 hours
AND table02.[Column_CreatedUTC] = COALESCE(@CreatedUTC, table02.[Column_CreatedUTC])
AND table03.[Column_EndDateUTC] = COALESCE(@BackupUTC, table03.[Column_EndDateUTC])

GROUP BY table03.Column_EndDateUTC, table02.Column_CreatedUTC
ORDER BY table02.Column_CreatedUTC ASC, DaysActive, table03.Column_EndDateUTC DESC

输出如下：

[DeviceCreated]..[DaysActive]..[LastCompleteBackup]
---------------------------------------------------------
[2/13/12 16:04]..[463 Days]....[5/21/13 12:14]
[2/12/13 22:37]..[97 Days].....[5/20/13 22:10]
于 2013-05-22T00:11:09.493 回答
0 
declare @StartDate datetime,
@EndDate datetime
set @StartDate = '10/01/2012 08:40:18.000'
set @EndDate =   '10/01/2012 09:52:48.000'

SELECT CONVERT(CHAR(8), CAST(CONVERT(varchar(23),@EndDate,121) AS DATETIME)
-CAST(CONVERT(varchar(23),@StartDate,121)AS DATETIME),8) AS TimeDiff
于 2013-10-31T18:41:17.267 回答
0

这是一个脚本写入副本，然后写入您的脚本文件并更改您的请求字段并退出

DECLARE @Sdate DATETIME, @Edate DATETIME, @Timediff VARCHAR(100)
SELECT @Sdate = '02/12/2014 08:40:18.000',@Edate='02/13/2014 09:52:48.000'
SET @Timediff=DATEDIFF(s, @Sdate, @Edate)
SELECT CONVERT(VARCHAR(5),@Timediff/3600)+':'+convert(varchar(5),@Timediff%3600/60)+':'+convert(varchar(5),@Timediff%60) AS TimeDiff
于 2014-03-06T10:36:42.810 回答
0 
DECLARE  @StartDate datetime = '10/01/2012 08:40:18.000'
        ,@EndDate   datetime = '10/10/2012 09:52:48.000'
        ,@DaysDifferent int = 0
        ,@Sec BIGINT

select @Sec = DateDiff(s, @StartDate, @EndDate)

IF (DATEDIFF(day, @StartDate, @EndDate) > 0)
    BEGIN
        select @DaysDifferent = DATEDIFF(day, @StartDate, @EndDate)
        select @Sec = @Sec - ( @DaysDifferent * 86400 )
        SELECT LTRIM(STR(@DaysDifferent,3)) +'d '+ LTRIM(STR(@Sec/3600, 5)) + ':' + RIGHT('0' + LTRIM(@Sec%3600/60), 2) + ':' + RIGHT('0' + LTRIM(@Sec%60), 2) AS [dd hh:mm:ss]
    END
ELSE
    BEGIN
        SELECT LTRIM(STR(@DaysDifferent,3)) +'d '+ LTRIM(STR(@Sec/3600, 5)) + ':' + RIGHT('0' + LTRIM(@Sec%3600/60), 2) + ':' + RIGHT('0' + LTRIM(@Sec%60), 2) AS [dd hh:mm:ss]
    END

----------------------------------------------------------------------------------
dd HH:MM:SS
9d 1:12:30
于 2014-03-26T14:07:18.357 回答
-1 
declare @StartDate datetime;
declare @EndDate datetime;
select @StartDate = '10/01/2012 08:40:18.000';
select @EndDate='10/04/2012 09:52:48.000';
select  cast(datediff(hour,@StartDate,@EndDate) as varchar(10)) + left(right(cast(cast(cast((@EndDate-@StartDate) as datetime) as time) as varchar(16)),14),6)
于 2017-04-27T01:23:28.740 回答