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ISO week number in CMD

How can I determine the (ISO 8601) week number in a Windows batch file?

Unfortunately WMIC PATH WIN32_LOCALTIME GET /FORMAT:LIST only has WeekInMonth...

I have found some very complex solutions. Is there no easier way?

like image 284
Michel de Ruiter Avatar asked Sep 07 '16 15:09

Michel de Ruiter


1 Answers

You can use Ritchie Lawrences's Date Functions. It is maintained on Gitub. https://ritchielawrence.github.io/batchfunctionlibrary/

Here is the code.

 ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: 
 :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
like image 97
Squashman Avatar answered Sep 19 '22 15:09

Squashman