有什么办法可以消除datetime
sql server中两者之间的区别?
例如,我的日期是
2010-01-22 15:29:55.090
2010-01-22 15:30:09.153
因此,结果应为14.063 seconds
。
有什么办法可以消除datetime
sql server中两者之间的区别?
例如,我的日期是
2010-01-22 15:29:55.090
2010-01-22 15:30:09.153
因此,结果应为14.063 seconds
。
Answers:
只是要增加有关DateDiff的警告,它计算您通过指定为单位的边界的次数,因此,如果您要查找精确的时间跨度,则会遇到问题。例如
select datediff (m, '20100131', '20100201')
给出的答案为1,因为它跨越了1月到2月的边界,因此即使跨度为2天,datediff也将返回值1-它跨越了1个日期边界。
select datediff(mi, '2010-01-22 15:29:55.090' , '2010-01-22 15:30:09.153')
再次给定值1,则它一次通过分钟边界,因此即使大约14秒,当使用Minutes作为单位时,它也将作为一分钟返回。
我可以提到MS SQL Server的四个非常有用的重要功能:
1)函数DATEDIFF()负责计算两个日期之间的差异,结果可能是在第一个参数(datepart)上指定的“年季度月日日日星期几小时分秒毫秒微秒纳秒” :
select datediff(day,'1997-10-07','2011-09-11')
2)您可以使用函数GETDATE()获取实际时间,并计算某些日期与实际日期之间的差值:
select datediff(day,'1997-10-07', getdate() )
3)另一个重要的函数是DATEADD(),用于使用datediff的相同datepart在datetime中转换某些值,您可以在一个基准日期中添加(带有正值)或减去(带有负值):
select DATEADD(day, 45, getdate()) -- actual datetime adding 45 days
select DATEADD( s,-638, getdate()) -- actual datetime subtracting 10 minutes and 38 seconds
4)函数CONVERT()可以根据需要格式化日期,它不是参数函数,但是您可以使用部分结果来格式化所需的结果:
select convert( char(8), getdate() , 8) -- part hh:mm:ss of actual datetime
select convert( varchar, getdate() , 112) -- yyyymmdd
select convert( char(10), getdate() , 20) -- yyyy-mm-dd limited by 10 characters
DATETIME cold以秒为单位计算,将这四个功能混合在一起的有趣结果是显示两个日期之间的格式差异um小时,分钟和秒(hh:mm:ss):
declare @date1 datetime, @date2 datetime
set @date1=DATEADD(s,-638,getdate())
set @date2=GETDATE()
select convert(char(8),dateadd(s,datediff(s,@date1,@date2),'1900-1-1'),8)
...结果为00:10:38(638s = 600s + 38s = 10分38秒)
另一个例子:
select distinct convert(char(8),dateadd(s,datediff(s, CRDATE , GETDATE() ),'1900-1-1'),8) from sysobjects order by 1
我尝试过这种方法,它奏效了。我使用了SQL Server 2016版
SELECT DATEDIFF(MILLISECOND,'2010-01-22 15:29:55.090', '2010-01-22 15:30:09.153')/1000.00;
不同的DATEDIFF函数是:
SELECT DATEDIFF(year, '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000');
SELECT DATEDIFF(quarter, '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000');
SELECT DATEDIFF(month, '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000');
SELECT DATEDIFF(dayofyear, '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000');
SELECT DATEDIFF(day, '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000');
SELECT DATEDIFF(week, '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000');
SELECT DATEDIFF(hour, '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000');
SELECT DATEDIFF(minute, '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000');
SELECT DATEDIFF(second, '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000');
SELECT DATEDIFF(millisecond, '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000');
参考:https : //docs.microsoft.com/zh-cn/sql/t-sql/functions/datediff-transact-sql? view = sql-server- 2017
好吧,我们都知道答案涉及DATEDIFF()
。但这只会给您带来一半的结果。如果要以人类可读的格式(两个DATETIME
值之间的分钟和秒)获取结果怎么办?
的CONVERT()
,DATEADD()
当然DATEDIFF()
功能是完美的一个更易读的结果,你的客户端可以使用,而不是数量。
即
CONVERT(varchar(5), DATEADD(minute, DATEDIFF(MINUTE, date1, date2), 0), 114)
这将为您提供以下信息:
高度:毫米
如果要提高精度,只需增加VARCHAR()
。
CONVERT(varchar(12), DATEADD(minute, DATEDIFF(MINUTE, date1, date2), 0), 114)
硬件:MM.SS.MS
有多种方法可以查看日期差,并且在比较日期/时间时有更多方法。这是我用来获取格式为“ HH:MM:SS”的两个日期之间的差的方法:
ElapsedTime AS
RIGHT('0' + CAST(DATEDIFF(S, StartDate, EndDate) / 3600 AS VARCHAR(2)), 2) + ':'
+ RIGHT('0' + CAST(DATEDIFF(S, StartDate, EndDate) % 3600 / 60 AS VARCHAR(2)), 2) + ':'
+ RIGHT('0' + CAST(DATEDIFF(S, StartDate, EndDate) % 60 AS VARCHAR(2)), 2)
我将其用于计算列,但您可以轻松地将其重写为UDF或查询计算。注意,此逻辑舍入小数秒;00:00.00到00:00.999被视为零秒,并显示为“ 00:00:00”。
如果您预计周期可能会超过几天,请在需要时将此代码切换为D:HH:MM:SS格式:
ElapsedTime AS
CASE WHEN DATEDIFF(S, StartDate, EndDate) >= 359999
THEN
CAST(DATEDIFF(S, StartDate, EndDate) / 86400 AS VARCHAR(7)) + ':'
+ RIGHT('0' + CAST(DATEDIFF(S, StartDate, EndDate) % 86400 / 3600 AS VARCHAR(2)), 2) + ':'
+ RIGHT('0' + CAST(DATEDIFF(S, StartDate, EndDate) % 3600 / 60 AS VARCHAR(2)), 2) + ':'
+ RIGHT('0' + CAST(DATEDIFF(S, StartDate, EndDate) % 60 AS VARCHAR(2)), 2)
ELSE
RIGHT('0' + CAST(DATEDIFF(S, StartDate, EndDate) / 3600 AS VARCHAR(2)), 2) + ':'
+ RIGHT('0' + CAST(DATEDIFF(S, StartDate, EndDate) % 3600 / 60 AS VARCHAR(2)), 2) + ':'
+ RIGHT('0' + CAST(DATEDIFF(S, StartDate, EndDate) % 60 AS VARCHAR(2)), 2)
END
以下查询应提供您正在寻找的确切内容。
select datediff(second, '2010-01-22 15:29:55.090' , '2010-01-22 15:30:09.153')
这是MSDN的链接,它提供有关datediff函数的全部功能。 https://msdn.microsoft.com/zh-CN/library/ms189794.aspx
SELECT DATEDIFF(yyyy, '2011/08/25', '2017/08/25') AS DateDiff
它为您提供年份中两个日期之间的差额
结果(2017-2011)= 6
句法:
DATEDIFF(interval, date1, date2)
因此,这不是我的答案,但我只是在网上搜索类似问题时发现了这个问题。这个家伙设置了一个程序来计算小时,分钟和秒。该链接和代码:
--Creating Function
If OBJECT_ID('UFN_HourMinuteSecond') Is Not Null
Drop Function dbo.UFN_HourMinuteSecond
Go
Exec(
'Create Function dbo.UFN_HourMinuteSecond
(
@StartDateTime DateTime,
@EndDateTime DateTime
) Returns Varchar(10)
As
Begin
Declare @Seconds Int,
@Minute Int,
@Hour Int,
@Elapsed Varchar(10)
Select @Seconds = ABS(DateDiff(SECOND ,@StartDateTime,@EndDateTime))
If @Seconds >= 60
Begin
select @Minute = @Seconds/60
select @Seconds = @Seconds%60
If @Minute >= 60
begin
select @hour = @Minute/60
select @Minute = @Minute%60
end
Else
Goto Final
End
Final:
Select @Hour = Isnull(@Hour,0), @Minute = IsNull(@Minute,0), @Seconds = IsNull(@Seconds,0)
select @Elapsed = Cast(@Hour as Varchar) + '':'' + Cast(@Minute as Varchar) + '':'' + Cast(@Seconds as Varchar)
Return (@Elapsed)
End'
)
PRINT DATEDIFF(second,'2010-01-22 15:29:55.090','2010-01-22 15:30:09.153')
select
datediff(millisecond,'2010-01-22 15:29:55.090','2010-01-22 15:30:09.153') / 1000.0 as Secs
result:
Secs
14.063
只是以为我会提到它。
CREATE FUNCTION getDateDiffHours(@fdate AS datetime,@ tdate as datetime)返回varchar(50)BEGIN DECLARE @cnt int DECLARE @cntDate datetime DECLARE @dayDiff int DECLARE @dayDiffWk int DECLARE @hrsDiff decimal(18)
DECLARE @markerFDate datetime
DECLARE @markerTDate datetime
DECLARE @fTime int
DECLARE @tTime int
DECLARE @nfTime varchar(8)
DECLARE @ntTime varchar(8)
DECLARE @nfdate datetime
DECLARE @ntdate datetime
-------------------------------------
--DECLARE @fdate datetime
--DECLARE @tdate datetime
--SET @fdate = '2005-04-18 00:00:00.000'
--SET @tdate = '2005-08-26 15:06:07.030'
-------------------------------------
DECLARE @tempdate datetime
--setting weekends
SET @fdate = dbo.getVDate(@fdate)
SET @tdate = dbo.getVDate(@tdate)
--RETURN @fdate
SET @fTime = datepart(hh,@fdate)
SET @tTime = datepart(hh,@tdate)
--RETURN @fTime
if datediff(hour,@fdate, @tdate) <= 9
RETURN(convert(varchar(50),0) + ' Days ' + convert(varchar(50),datediff(hour,@fdate, @tdate))) + ' Hours'
else
--setting working hours
SET @nfTime = dbo.getV00(convert(varchar(2),datepart(hh,@fdate))) + ':' +dbo.getV00(convert(varchar(2),datepart(mi,@fdate))) + ':'+ dbo.getV00(convert(varchar(2),datepart(ss,@fdate)))
SET @ntTime = dbo.getV00(convert(varchar(2),datepart(hh,@tdate))) + ':' +dbo.getV00(convert(varchar(2),datepart(mi,@tdate))) + ':'+ dbo.getV00(convert(varchar(2),datepart(ss,@tdate)))
IF @fTime > 17
begin
set @nfTime = '17:00:00'
end
else
begin
IF @fTime < 8
set @nfTime = '08:00:00'
end
IF @tTime > 17
begin
set @ntTime = '17:00:00'
end
else
begin
IF @tTime < 8
set @ntTime = '08:00:00'
end
-- used for working out whole days
SET @nfdate = dateadd(day,1,@fdate)
SET @ntdate = @tdate
SET @nfdate = convert(varchar,datepart(yyyy,@nfdate)) + '-' + convert(varchar,datepart(mm,@nfdate)) + '-' + convert(varchar,datepart(dd,@nfdate))
SET @ntdate = convert(varchar,datepart(yyyy,@ntdate)) + '-' + convert(varchar,datepart(mm,@ntdate)) + '-' + convert(varchar,datepart(dd,@ntdate))
SET @cnt = 0
SET @dayDiff = 0
SET @cntDate = @nfdate
SET @dayDiffWk = convert(decimal(18,2),@ntdate-@nfdate)
--select @nfdate,@ntdate
WHILE @cnt < @dayDiffWk
BEGIN
IF (NOT DATENAME(dw, @cntDate) = 'Saturday') AND (NOT DATENAME(dw, @cntDate) = 'Sunday')
BEGIN
SET @dayDiff = @dayDiff + 1
END
SET @cntDate = dateadd(day,1,@cntDate)
SET @cnt = @cnt + 1
END
--SET @dayDiff = convert(decimal(18,2),@ntdate-@nfdate) --datediff(day,@nfdate,@ntdate)
--SELECT @dayDiff
set @fdate = convert(varchar,datepart(yyyy,@fdate)) + '-' + convert(varchar,datepart(mm,@fdate)) + '-' + convert(varchar,datepart(dd,@fdate)) + ' ' + @nfTime
set @tdate = convert(varchar,datepart(yyyy,@tdate)) + '-' + convert(varchar,datepart(mm,@tdate)) + '-' + convert(varchar,datepart(dd,@tdate)) + ' ' + @ntTime
set @markerFDate = convert(varchar,datepart(yyyy,@fdate)) + '-' + convert(varchar,datepart(mm,@fdate)) + '-' + convert(varchar,datepart(dd,@fdate)) + ' ' + '17:00:00'
set @markerTDate = convert(varchar,datepart(yyyy,@tdate)) + '-' + convert(varchar,datepart(mm,@tdate)) + '-' + convert(varchar,datepart(dd,@tdate)) + ' ' + '08:00:00'
--select @fdate,@tdate
--select @markerFDate,@markerTDate
set @hrsDiff = convert(decimal(18,2),datediff(hh,@fdate,@markerFDate))
--select @hrsDiff
set @hrsDiff = @hrsDiff + convert(int,datediff(hh,@markerTDate,@tdate))
--select @fdate,@tdate
IF convert(varchar,datepart(yyyy,@fdate)) + '-' + convert(varchar,datepart(mm,@fdate)) + '-' + convert(varchar,datepart(dd,@fdate)) = convert(varchar,datepart(yyyy,@tdate)) + '-' + convert(varchar,datepart(mm,@tdate)) + '-' + convert(varchar,datepart(dd,@tdate))
BEGIN
--SET @hrsDiff = @hrsDiff - 9
Set @hrsdiff = datediff(hour,@fdate,@tdate)
END
--select FLOOR((@hrsDiff / 9))
IF (@hrsDiff / 9) > 0
BEGIN
SET @dayDiff = @dayDiff + FLOOR(@hrsDiff / 9)
SET @hrsDiff = @hrsDiff - FLOOR(@hrsDiff / 9)*9
END
--select convert(varchar(50),@dayDiff) + ' Days ' + convert(varchar(50),@hrsDiff) + ' Hours'
RETURN(convert(varchar(50),@dayDiff) + ' Days ' + convert(varchar(50),@hrsDiff)) + ' Hours'
结束
Sol-1:
select
StartTime
, EndTime
, CONVERT(NVARCHAR,(EndTime-StartTime), 108) as TimeDiff
from
[YourTable]
Sol-2:
select
StartTime
, EndTime
, DATEDIFF(hh, StartTime, EndTime)
, DATEDIFF(mi, StartTime, EndTime) % 60
from
[YourTable]
Sol-3:
select
DATEPART(hour,[EndTime]-[StartTime])
, DATEPART(minute,[EndTime]-[StartTime])
from
[YourTable]
Datepart效果最好
datediff
答案,但似乎没有一个提醒您,根据参数顺序,您可能会从中得到否定的结果。