Regarding leap years, for anyone who’s interested, I recommend going beyond the simple divide by 4 logic, as it will not work in the year 2100. Yes, few of us here today will be around then, but similar logic will likely still be in use in whatever form it exists then. I know there will be plenty of false leap year determinations in 2100.

The additional rule is that for years ending in 00, they must be evenly divisible for 400 in order to be a leap year. Thus, 2000 was a leap year, but 1900 was not, and 2100, 2200, and 2300 will not be.

without the coding, in easy words:
date (01 + current month + current year) – 1 day

d wrkdate         s               d                    
 /free                                                 
      wrkdate = %date() - %days(%subdt(%date():*days));
      dump(a);                                         
      *inlr = *on;                                     
 /end-free                                             

otherwise, replace %date() with the date you want to find the previous month-end date for.

Kevin C. Ketzler – Affiliated

That first “C LastMonth WhenEq 1″ should be “C LastMonth WhenEq 2″, but then you guys already knew that, right.

Regards,

Martin.

The last day of the month is in essence a constant dependent on which month it is and whether or not it’s a leap year, so you could hard-code for this :-

     C     CurrentMonth  Sub       1             LastMonth         2 0
     C                   Select                                       
     C     LastMonth     WhenEq    1                                  
     C     Year          Div       4             Dummy             5 0
     C                   Mvr                     LeapYear          1 0
     C     LeapYear      IfNe      *Zeros                             
     C                   Z-add     28            DayNumber            
     C                   Else                                         
     C                   Z-add     29            DayNumber            
     C                   Endif                                        
     C     LastMonth     WhenEq    1                                  
     C     LastMonth     Oreq      3                                  
     C     LastMonth     Oreq      5                                  
     C     LastMonth     Oreq      7                                  
     C     LastMonth     Oreq      8                                  
     C     LastMonth     Oreq      10                                 
     C     LastMonth     Oreq      12                                 
     C                   Z-add     31            DayNumber            
     C     LastMonth     WhenEq    4                                  
     C     LastMonth     Oreq      6                                  
     C     LastMonth     Oreq      9                                  
     C     LastMonth     Oreq      11                                 
     C                   Z-add     30            DayNumber            
     C                   EndSl      

This isn’t a perfect solution (if you can divide the year by 4, it’s not *always* a leap year), but it should work for most situations.

Regards,

Martin Gilbert.

as you can see – manuipulate the date in the data structure so that the current date is changed to the first of the current month – then subtract 1 day

DTODAY DS
D TODAYD D

/FREE
TODAYD = %DATE();
%SUBST(TODAY:9:2) = ’01′;
TODAYD-=%DAYS(1);
*INLR = *ON;
/END-FREE