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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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