-- Find Hours, Minutes and Seconds in between two datetime DECLARE@Firstdatetime DECLARE@Seconddatetime SET@First='04/02/2008 05:23:22' SET@Second= getdate() SELECTDATEDIFF(day,@First,@Second)*24asTotalHours, DATEDIFF(day,@First,@Second)*24*60asTotalMinutes, DATEDIFF(day,@First,@Second)*...
DateDif StartDate EndDate Days Hours Minutes Seconds MS --- --- --- --- --- --- --- --- 1900-06-02 10:40:07.913 2009-02-13 02:44:37.923 2009-07-15 13:24:45.837 152 10 40 7 913 (1 row(s) affected) Share Improve this answer Follow edited Jan 31, 2016 at 20:55 ...
| Hours | DATEDIFF(hh, start, end) | days_diff * 24 + DATE_PART('hour', end - start ) | | Minutes | DATEDIFF(mi, start, end) | hours_diff * 60 + DATE_PART('minute', end - start ) | | Seconds | DATEDIFF(ss, start, end) | minutes_diff * 60 + DATE_PART('minute', ...
SQL Server DATEDIFF function returns the difference in seconds, minutes, hours, days, weeks, months, quarters and years between 2 datetime values. Quick Example: -- The difference is days between today and yesterday SELECT DATEDIFF(dd, GETDATE(
118 years, 11 months, 11 days, 7 hours, 8 minutes and 1.123 seconds 範例:Azure Synapse Analytics 和 Analytics Platform System (PDW) 這些範例會使用不同的運算式類型,當作startdate和enddate參數的引數。 J. 指定 startdate 和 enddate 的數據行 ...
118 years, 11 months, 11 days, 7 hours, 8 minutes and 1.123 seconds 範例:Azure Synapse Analytics 和 Analytics Platform System (PDW) 這些範例會使用不同的運算式類型,當作startdate和enddate參數的引數。 J. 指定 startdate 和 enddate 的數據行 ...
Days, A, TRUNC(A*24) Hours, TRUNC(A*24*60 - 60*TRUNC(A*24)) Minutes, TRUNC(A*24*60*60 - 60*TRUNC(A*24*60)) Seconds, TRUNC(A*24*60*60*100 - 100*TRUNC(A*24*60*60)) mSeconds from ( select trunc(sysdate) Days,
declare @days as int, @hours as int, @minutes as int, @seconds as int; set @seconds = datediff( second, @dt2, @dt1); set @days = @seconds / (24 * 60 * 60) set @seconds = @seconds - @days * 24 * 60 * 60 set @hours = @seconds / (60 * 60); ...
set @offset=datediff(day,'1900-01-30',@solarday) --确定农历年开始 set @i=1900 --set @offset=@soldata while @i<2050 and @offset>0 begin set @ydays=348 set @mleapnum=0 select @ilunar=dataint from solardata where yearid=@i ...
See https://learn.microsoft.com/en-us/sql/t-sql/functions/datediff-transact-sql?view=sql-server-ver15#return-value For millisecond, the maximum difference between startdate and enddate is 24 days, 20 hours, 31 minutes and 23.647 seconds. If you need millisecond above that level...