Michael Alipio am Mittwoch, 24. Januar 2007 04:21:
> From: John W. Krahn <[EMAIL PROTECTED]>
> Sent: Wednesday, January 24, 2007 10:57:51 AM
> > Yes, the substitution operator (s///) returns true (1) or false ('') in
> > either list or scalar context. To do want you want you have to do the
> > assignment first and then do the substitution:
> >
> > my $newname = $_;
> > $newname =~ s/^\w+-//;
> >
> > Or in one statement:
> >
> > ( my $newname = $_ ) =~ s/^\w+-//;
>
> I've already figured that one out. However, I want to use variables for my
> regexp pattern. So I can replace "axis" with whatever I my first program
> argument is.
[...]
Hi Michael
> find (\&renamefiles, './');
> my $name = shift;
You initialize $name after the call to find(), so renamefiles() has nothing in
$name. Switch these lines (and test the user provided contents of $name)
> sub renamefiles{
> if ($_ =~ /$name/){
if ($_ =~ /\Q$name\E/){
just in case $name contains chars that are special to the regex engine.
> my $oldname = $_;
> $_ =~ s/\w+-//;
> #rename ($oldname, $_)
> print "$oldname will be renamed to $_\n";
> }
> }
>
> I got many of this:
>
> Use of uninitialized value in regexp compilation at test.pl line 11.
Dani
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