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(
(s, -@seconds, @date2) SELECT @milliseconds=DATEDIFF(ms, @date1, @date2) SELECT @result= ISNULL(CAST(NULLIF(@years,0) AS VARCHAR(10)) + ' years,','') + ISNULL(' ' + CAST(NULLIF(@months,0) AS VARCHAR(10)) + ' months,','') + ISNULL(' ' + CAST(NULLIF(@days,0) AS...
| Seconds | DATEDIFF(ss, start, end) | minutes_diff * 60 + DATE_PART('minute', end - start ) | PostgreSQL-年中的日期差异 考虑使用 SQL Server 函数来计算以年为单位的两个日期之间的差: SQL Server: --DifferencebetweenOct02,2011andJan01,2012inyearsSELECTDATEDIFF(year,'2011-10-02','2012...
DATEDIFF( microsecond,@date1,@date2)ASMicroseconds, DATEDIFF( nanosecond,@date1,@date2)ASNanoseconds; AI代码助手复制代码 结果: The datedifffunctionresultedinan overflow. The numberofdateparts separating twodate/time instancesistoo large.Trytouse datediffwitha less precise datepart. AI代码助手复制代码...
DATEDIFF(day, DATEADD(year, DATEDIFF(year, startdate, enddate), startdate), enddate) 这个方法的作用是先计算两个日期之间的年份差异,然后将startdate的年份加上这个差异,得到一个新的日期,最后计算这个新日期和enddate之间的天数差异。这样就可以忽略年份,只计算日期部分的差异。 例如,如果startdate为'2021-...
SELECT@seconds=DATEDIFF(s, @date1, @date2); IF DATEADD(s, -@seconds, @date2) < @date1SELECT@seconds= @seconds-1;SET@date2 =DATEADD(s, -@seconds, @date2);SELECT@milliseconds =DATEDIFF(ms, @date1, @date2);SELECT@result=ISNULL(CAST(NULLIF(@years,0)ASVARCHAR(10)) +' years,',...
it works. It seams that if a boundary to the next minute is crossed the value ofDATEDIFF(SECONDS...)always yields 1 while for the same interval all within one minute the seconds yield 0. Please have a look at rows with id 8 and 9 as well as 10 and 11 in the ...
],DATEDIFF(MONTH,@statetdate,@enddate)AS[Months],DATEDIFF(WEEK,@statetdate,@enddate)AS[Weeks],DATEDIFF(DAY,@statetdate,@enddate)AS[Days],DATEDIFF(HOUR,@statetdate,@enddate)AS[Hours],DATEDIFF(MINUTE,@statetdate,@enddate)AS[Minutes],DATEDIFF(SECOND,@statetdate,@enddate)AS[Seconds]...
Datediff return an integer, not a float or numeric. To get what you want, perhaps you should do the diff in seconds and then divide: select cast(datediff(second, min(start_time), min(complete_time))/60.0 as decimal (10,10)) from TRACKED_OBJECT_HISTORY TOH1 where TOH1.op_name = '...
TIMEFROMPARTSTIMEFROMPARTS ( hour, minute, seconds, fractions, precision )对指定的时间返回 time 值(具有指定精度)。time( precision )Deterministic 返回日期和时间差异值的函数 展开表 函数语法返回值返回数据类型确定性 DATEDIFFDATEDIFF ( datepart, startdate, enddate )返回两个指定日期之间所跨的日期或时间 ...