• Subject: RE: last day of month (was MIDRANGE-L Digest V2 #1546)
  • From: Jon Erickson <jerickson@xxxxxxx>
  • Date: Thu, 14 Oct 1999 09:16:48 -0700

Barbara,

Correct me if I'm wrong, but I believe by using AddDur to a date with Day=1,
(e.g. 1999-02-01), you'll receive the 1st day of the next month,
(1999-03-01). Therefore, by initializing an *ISO date field, the result is
'0001-01-01' thus, when you overlay the input &year and &month, do the
AddDur 1 month, Subdur 1 day, you'll always get the last day of the &year
and &month, such as the case of the code following.

d                 DS                                    
d Isodate                         d   Inz               
d  Isoyyyy                       4  0 Overlay(Isodate)  
d  Isomm                         2  0 Overlay(Isodate:6)
 
c                   Eval      Isoyyyy = Pyear
c                   Eval      Isomm = Pmonth
Add 1 month via ADDDUR
c                   AddDur    1:*Months     Isodate

Subtract 1 day via SUBDUR
c                   SubDur    1:*Days       Isodate
 

Regards,
Jon A. Erickson
Sr. Programmer Analyst
800.COM Inc.
1516 NW Thurman St
Portland, OR  97209-2517

Direct: 503.944.3613
Fax: 503.944.3690
Web: http://800.com


-----Original Message-----
From: bmorris@ca.ibm.com [mailto:bmorris@ca.ibm.com]
Sent: Wednesday, October 13, 1999 4:32 PM
To: MIDRANGE-L@midrange.com
Subject: Re: last day of month (was MIDRANGE-L Digest V2 #1546)





>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
in
>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
9999:

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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