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Rob said:
> The fix to Marvin's solution would be the following:
> parms.transNo = (%editc(%dec(transNo:(%size(transNo):0) + 1):'X'));
> Notice the left before the %size, and the right at the end?

This will still give an 11a result, not the 10a he desires.

I am assuming he figures it will never overflow past 1 billion since he's 
not accounting for that eventuality, so if he can ALSO assume it will 
never overflow past 999 mil, this will work for him:

parms.transNo = 
(%editc(%dec(%subst(transNo:2:%size(transno-1)):(%size(transNo)-1:0) + 
1):'X'));

  or 

parms.transNo = %editc(
                       (1+%dec(
                            %subst(transNo:2:%size(transno-1)):
                            (%size(transNo)-1):
                             0) 
                       ):
                       'X');


but now we're getting to the point of diminishing returns - way too 
complicated.


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