RE: Removing a dash from a phone number

I'd rather use a check for ' ' then use %replace on the non-blanks or spin 
through that puppy with a DO While (for next anyone) and perform a %replace on 
the string.. so like this..

c                   For       x=1 to %Len(%Trim(string))

c                   Select
c                   When      %Subst(string:x:1) <> '_'  and FirstSpace
 * Reset First Space for next block of alpha's
c                   Eval        Firstspace = *On
c                   iter
c                   When      %Subst(string:x:1) = '_'  and FirstSpace
* Set FirstSpace and skip the first one to make sure you still have one space 
between text
c                   Eval        Firstspace = *Off
c                   iter
c                   When      %Subst(string:x:1) = '_'  and not FirstSpace
 * Replace it out with NOTHING (compress...)
c                   Eval      Stinin = %replace(' ':String:x:1)
c                   iter
c                   endsl
c                   endfor

Replace the '_' with a space - it just didn't look good while entering...

Andrew Borts / Webmaster
Seta Corporation
6400 East Rogers Circle
Boca Raton, FL 33499

E-mail: Andrewb@setacorporation.com
Corporate web site http://www.setacorporation.com
E-Commerce web site http://www.palmbeachjewelry.com
                              http://www.myfreeitems.com
Voice: 561-994-2660 Ext. 2211 / Fax: 561-997-0774

-----Original Message-----
From: Amitava Banerjee [mailto:Amitava.Banerjee@trin.net]
Sent: Tuesday, January 22, 2002 2:48 PM
To: rpg400-l@midrange.com
Subject: Removing a dash from a phone number

Suppose the field name for your phone number is OLDPH, then the code might be 
like this

                   Z-ADD         1                I               10
                  MOVE         *BLANKS    NEWPH   8
I                 DOWLE       8
1                SUBST        OLDPH:I    CH           1
CH             IFNE            '-'
NEWPH    CAT             CH:0           NEWPH
                  ENDIF
                  ADD             1                 I
                  ENDDO

In ILE it will be


D pos                s              1s 0 inz(1)
D newph           s              8a   inz(*blanks)
D oldph             s              8a   inz('123-4567')
D ch                 s              1a   inz(*blanks)
C                   DOW       POS <= 8
C                   EVAL      CH = %SUBST(OLDPH:POS:1)
C                   IF            ch <> '-'
C     newph   CAT        ch:0          newph
C                   endif
C                   eval      pos = pos + 1
C                   enddo


Hope it helps.
Amitava


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