CMD中的ISO周编号

Question

如何确定Windows批处理文件中的（ISO 8601）周数？

不幸的是WMIC PATH WIN32_LOCALTIME GET /FORMAT:LIST只有WeekInMonth ...

我发现了一些非常 复杂的 解决方案。有没有更简单的方法？

Answer 1

您可以使用Ritchie Lawrences的日期函数。它在Gitub上维护。 https://ritchielawrence.github.io/batchfunctionlibrary/

这是代码。

 ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: 
 :DateToWeek %yy% %mm% %dd% yn cw dw
 ::
 :: By:   Ritchie Lawrence, Updated 2002-11-20. Version 1.1
 ::
 :: Func: Returns an ISO 8601 Week date from a calendar date.
 ::       For NT4/2000/XP/2003.
 :: 
 :: Args: %1 year component to be converted, 2 or 4 digits (by val)
 ::       %2 month component to be converted, leading zero ok (by val)
 ::       %3 day of month to be converted, leading zero ok (by val)
 ::       %4 var to receive year, 4 digits (by ref)
 ::       %5 var to receive calendar week, 2 digits, 01 to 53 (by ref)
 ::       %6 var to receive day of week, 1 digit, 1 to 7 (by ref)
 :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
 setlocal ENABLEEXTENSIONS
 set yy=%1&set mm=%2&set dd=%3
 if 1%yy% LSS 200 if 1%yy% LSS 170 (set yy=20%yy%) else (set yy=19%yy%)
 set /a dd=100%dd%%%100,mm=100%mm%%%100
 set /a z=14-mm,z/=12,y=yy+4800-z,m=mm+12*z-3,Jd=153*m+2
 set /a Jd=Jd/5+dd+y*365+y/4-y/100+y/400-32045
 set /a y=yy+4798,Jp=y*365+y/4-y/100+y/400-31738,t=Jp+3,Jp=t-t%%7
 set /a y=yy+4799,Jt=y*365+y/4-y/100+y/400-31738,t=Jt+3,Jt=t-t%%7
 set /a y=yy+4800,Jn=y*365+y/4-y/100+y/400-31738,t=Jn+3,Jn=t-t%%7
 set /a Jr=%Jp%,yn=yy-1,yn+=Jd/Jt,yn+=Jd/Jn
 if %Jd% GEQ %Jn% (set /a Jr=%Jn%) else (if %Jd% GEQ %Jt% set /a Jr=%Jt%)
 set /a diff=Jd-Jr,cw=diff/7+1,wd=diff%%7,wd+=1
 if %cw% LSS 10 set cw=0%cw%
 endlocal&set %4=%yn%&set %5=%cw%&set %6=%wd%&goto :EOF