Re: trimming non-numerics

["mlazarus@xxxxxxxx" <mlazarus@xxxxxxxx>]

Hans,

 Why do you say the using %Xlate / %Replace is overkill?  Is it performance
you're concerned with?  It would be fewer statements (either one or two)
and doesn't reduce legibility, IMHO.

 Obviously, if there are other, unexpected, characters to strip out of the
source string, then that's a different story.  But that did not seem to be
the case here.


 As an aside, the statement:

        if '0' <= %subst(source:i:1) and %subst(source:i:1) <= '9';
 should read:
        if '0' >= %subst(source:i:1) and %subst(source:i:1) <= '9';

 -mark

Original Message:
-----------------
From: Hans Boldt boldt@xxxxxxxxxx
Date: Tue, 01 Jun 2004 09:18:07 -0400
To: rpg400-l@xxxxxxxxxxxx
Subject: Re: trimming non-numerics


Jeffrey Lee wrote:
> I was wondering how I could strip any non numberic characters from a set
of numbers submitted via a html form. For example (504) 555-5555 would be
5045555555 or 504-555-5555 would be 5045555555   Like %trim does with
spaces is there something to deal with non-numberics?
> --

If you were programming in Python, you could do this with the simple 
expression (''.join([c for c in source if c.isdigit()])). ;-)

But since this is an RPG list, you probably want an RPG answer, right?

Others have suggested using some combination of the built-in functions 
%XLATE and %REPLACE. Personally, I thinks that's overkill. I'd recommend 
the obvious approach of simply looping over the source string selecting 
the digits out of the string (just like what the above Python expression 
does):

------------------------------------------------------------------
D i               s             10i 0
D source          s             50a   varying
D target          s             50a   varying
  /free
     source = 'jafgjh934hdfg94k3ghsd0jdfg0345jdfg';
     target = '';

     // Extract digits out of source string
     for i = 1 to %len(source);
        if '0' <= %subst(source:i:1) and %subst(source:i:1) <= '9';
           target += %subst(source:i:1);
        endif;
     endfor;

     // Display results
     dsply source;
     dsply target;
     *inlr = *on;
  /end-free
------------------------------------------------------------------

Oh yeah, if you're not on V5R2, replace the middle statement of the loop 
with:

           target = target + %subst(source:i:1);

Cheers! Hans


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