Grab a perl book or search the web for concatenating variables together...
Unfortunately the current place I am in doesn't have perl on the server for me
to try and play around more, but the gist is the same. Your problem arise from
setting it up as string during the concatination process.
Set the responses in an array and just walk the array for a quick and dirty
solution.
---- Bobby <[EMAIL PROTECTED]> wrote:
> Thanks for both of your suggestions. I've tried them both but still getting
> same result. Any other suggestions?
>
> Thanks.
>
> Wolf <[EMAIL PROTECTED]> wrote:
> ---- Bobby wrote:
> > Wolf,
> >
> > I still don't understand, so set my $strB = "$strA($count)"; ? That didn't
> > worked.
> >
> > Wolf wrote:
> > ---- Bobby wrote:
> > > Hi all,
> > >
> > > I'm trying to write a simple do until loop to print out the value of
> > > $strA0 through $striA3. What i'm doing is replacing the value of 0
> > > through 3 in the $strA by joining two strings (my $strB = "strA" .
> > > $count;). Right now my script is printing $strB as below. How do i get
> > > perl to print the value of $strA0 through $strA3 inside of my do until
> > > loop? i.e.:
> > >
> > > Desired Outcome:
> > > VarB: A0
> > > VarB: A1,b,c
> > > VarB: A 2
> > > VarB: A3,d,e
> > >
> > > Current Outcome:
> > > VarB:strA0
> > > VarB:strA1
> > > VarB:strA2
> > > VarB:strA3
> > >
> > >
> > > #!/usr/bin/perl
> > > use strict;
> > > use warnings;
> > >
> > > my $strA0="A0";
> > > my $strA1="A1,b,c";
> > > my $strA2="A2";
> > > my $strA3="A3,d,e";
> > >
> > > my $count = 0;
> > > until ($count == 4){
> > > my $strB = "strA" . $count;
> > > print "VarB:$strB\n";
> > > $count++;
> > > }#end until loop
> > >
> > > Thanks,
> > >
> > > Mike
> >
> > You can either turn them into an array and walk the array with your count
> >
> > You are setting $strB as the string strA instead of $strA($count) which is
> > where you are getting hosed up.
> >
> > My perl server is down or I'd play with the code a bit to get it the exact
> > answer, but that should point you in the right direction.
> >
> > HTH,
> > Wolf
> >
>
> No Bobby, more along the lines of:
> my $B = "strA" .$count;
> my $strB = $B;
> print "VarB:$strB\n";
>
> But you are going to have to keep playing with it if it doesn't work right.
>
> Wolf
>
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