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

Question: I'm curious how to interpolate a matched variable ($1) into the replacement variable of the regex. 
my $text    = "Other Text. From {foo bar foo bar}";
my $pattern = 'From \{(.*?)\}';
my $replace = 'To {$1}';
$text       =~ s/$pattern/$replace/ge;

# text should equal "Other Text. To {foo bar foo bar }";

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I was able to accomplish it by creating an external function and passing $1 as a parameter to that function, but I figure there's an easier way. Also, I was not able to create and call the inline function in one step.

Please no comments like, "why don't you $text=~s/From/To/"?


As always, I appreciate the help.

Demize




Downvotes are welcome, just please comment as to why you are doing them. Thank you.

Replies are listed 'Best First'.
Re: Pass Matched Variable into Regex Replace Variable
by Anonyrnous Monk (Hermit) on Dec 17, 2010 at 18:19 UTC

    You could do 

    ...
my $replace = '"To {$1}"';
$text =~ s/$pattern/$replace/gee;   # eval twice

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[d/l]
      I feel like I've tried that and it didn't work.

      perl -e '$t="begin{foobar}";$p="begin{(.*?)}";$r="\"end{\$1}\""; $t=~s/$p/$r/gee; print "$t\n\n"'
      #Output: end{foobar}

      Perhaps I'm doing something wrong? Edit:I didn't escape the dollar sign in the variable. The suggestion works. I'll apply it to my program and see if everything works :) Thank you!
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[d/l]
[select]
        $r="\"end{$1}\"";

        The double quotes interpolate the $1. Use $r=q("end{$1}") in this case.

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[d/l]
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Re: Pass Matched Variable into Regex Replace Variable
by AnomalousMonk (Archbishop) on Dec 17, 2010 at 21:27 UTC
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