|
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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