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

I'm writing a find/replace script, and am running into issues with the ampersand and pound signs.

q[<IL.Check/>]=>q[<&#x221a;>],

In the previous example, I'm trying to find <IL.Check/> and replace it with &#x221a;. I have even added quotes, as in "&#x221a;" or &"#"x221a;, but in both cases the quotes appear in the output.

Am I adding quotes in the wrong place, or do I also need to add something for the ampersand?

Any help is appreciated, thanks.

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Re: How to ignore ampersand and pound signs?
by bichonfrise74 (Vicar) on Nov 04, 2009 at 01:25 UTC
    Hmm... not sure if I understood your question, but hopefully this would help you out. 
    #!/usr/bin/perl

use strict;

my $a = qq{<IL.Check/>};
my $b = qq{<&#x221a;>};

print "Old word: $a\n";

$a =~ s/$a/$b/;
print "New word: $a\n";

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      Thank you!

      The double q (qq) worked!

      Sorry for the basic question, I'm just starting out with Perl. :-)

[reply]
        Nonsense. The single quote you originally posted works just as well. 
        my $s = q[<IL.Check/>];
my $r = q[<&#x221a;>];

print "Old word: $s\n";
$s =~ s/$s/$r/;
print "New word: $s\n";

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        There's no interpolation or escape sequences here. q vs qq is moot.

        The real problem is that you are using text as a regex pattern without first converting it to a regex pattern. ("#" is special under "x".) Use quotemeta or its alias \Q..\E: 

        my $s = q[<IL.Check/>];
my $r = q[<&#x221a;>];

print "Old word: $s\n";
$s =~ s/\Q$s\E/$r/;
print "New word: $s\n";

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