Re: The case of outnumbered numerator

On Wed, Jun 17, 2015 at 1:44 AM, Gad Miron <gadmiron@xxxxxxxxx> wrote:

We have a six digits numerator that approaches 899999.

Since it happens to be a CHAR field I would like to start using the
letters A-Z
in lexical order when incrementing this field so that incrementing
899999 will result in 9AAAAA , 9AAAAB , ......., 9AAAAZ, 9AAAA0, 9AAAA1 etc.
(This way the numerator will sort the EBCDIC natural way)

You are basically asking for base 36 arithmetic. (Our usual familiar
arithmetic is base 10, hexadecimal is base 16.) There are already
functions out there for this, though they tend to put 0..9 first and
use A as 10, B as 11, etc.

So look around (there are even some relevant posts in the midrange.com
archives). If you use one of the usual (starting with 0 instead of A)
algorithms, you can just wrap it with the necessary character mapping
at the beginning and end.

I don't have time to elaborate further at the moment. I can help more
later if needed.

John Y.