• Subject: Re: last day of month (was MIDRANGE-L Digest V2 #1546)
  • From: bmorris@xxxxxxxxxx
  • Date: Wed, 13 Oct 1999 19:32:08 -0400

>Date: Wed, 13 Oct 1999 14:55:34 -0400
>From: Jon.Paris@halinfo.it
>The following will supply the date of the last day of the month for the month
>which InputDate falls.
> D InputDate                       D   DATFMT(*ISO)
> D Temp                           3P 0
> C                   Extract   InputDate:*D  Temp
> C                   Eval      Temp = Temp - 1
> C                   AddDur    1:*M          InputDate
> C                   SubDur    1:*D          InputDate
>Thats's all there is to it!!

Not quite all ... your code works fine if InputDate is already at the first of
the month, although you could leave out the first two lines ... but

Even if you subdur'd temp instead of 1 on the last line, you'd be off by one day
- this gives the beginning of the next month.  Sometimes.  You also have to
extract the days AFTER adding one month (finding the end of month containing say
Jan 30 requires subtracting 28 or 29 days, not 30 days ...)

Try this - it works for every date up to but not including dates in December

D InputDate       s               D   DATFMT(*ISO)
D Temp            s              3P 0

C                   AddDur    1:*M          InputDate
C                   Extrct    InputDate:*D  Temp
C                   SubDur    Temp:*D       InputDate
C                   return    InputDate

Barbara Morris

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