ibanix has asked for the wisdom of the Perl Monks concerning the following question:

Hi fellow monks,

How do I do a substitute on a variable and assign it to a another variable, w/o actually editing the first variable?

This is what I've been trying 
my ($path) = $logfilepath =~ s/$ref_srv/$server/i;

[download]
but this returns "1" into $path; which is suppose is the return value of the s/$ref_srv/$server/i operation.

I'll be reading perldoc, but perhaps someone will have a quicker/better/shorter way to do this.

Thanks,
ibanix

$ echo '$0 & $0 &' > foo; chmod a+x foo; foo;

Replies are listed 'Best First'.
Re: s/// operator and binding
by Ovid (Cardinal) on Jan 28, 2003 at 00:25 UTC

    Personally, I feel the answer is rather ugly and I would avoid it. You may think it's okay, though.

    (my $path = $logfilepath) =~ s/$ref_srv/$server/i;

    If $ref_srv has characters which might have a special meaning in a regex and you wish to avoid this:

    (my $path = $logfilepath) =~ s/\Q$ref_srv\E/$server/i;

    Cheers,
    Ovid

    New address of my CGI Course.
    Silence is Evil (feel free to copy and distribute widely - note copyright text)

[reply]
[d/l]
[select]
      D'OH! That's so obvious!

      Thanks. Why do you find it ugly? Is there a better way to do it?

      Cheers,
      ibanix

      $ echo '$0 & $0 &' > foo; chmod a+x foo; foo;
[reply]
[d/l]

        I don't want to answer for Ovid but I especially dislike mixing the declaration with that kind of assignment. I'd prefer 

        my $foo; 
($foo = $bar) =~ s/baz/qux/;

        [download]
        but if you are going to separate them anyway, it makes perfect sense to do it in a way that makes all the steps blindingly obvious such as this: 
        my $foo = $bar; 
$bar =~ s/baz/qux/;

        [download]
        There's rarely a good reason to sacrifice clarity for brevity.

        -sauoq
"My two cents aren't worth a dime.";
[reply]
[d/l]
[select]
        Well it's ugly to me because it's not immediately apparent what you're trying to do. Why not just do this 
        my($path) = $logfilepath;
$path =~ s/$ref_srv/$server/i;

        [download]
        Choosing between the two, I'd go with the simpler. Just because you can do something doesn't mean you should ;)

        update (broquaint): changed <pre> to <code>

[reply]
[d/l]
[select]

        For me, this is an aesthetic thing that you can safely ignore. If I want extra parentheses, I'd code in Lisp :)

        If you like them, however, it's not a problem, so don't worry about my blather.

        Cheers,
        Ovid

        New address of my CGI Course.
        Silence is Evil (feel free to copy and distribute widely - note copyright text)

[reply]
Re: s/// operator and binding
by Aristotle (Chancellor) on Jan 28, 2003 at 01:40 UTC 
    s/$ref_srv/$server/i for my $path = $logfilepath;

    [download]

    Makeshifts last the longest.

[reply]
[d/l]

Back to Seekers of Perl Wisdom