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How Do I Use a regex back-reference in a Command Line Argument?

ferulebezel
New User

Nov 29, 2009, 1:15 AM

Post #1 of 2 (1728 views)

How Do I Use a regex back-reference in a Command Line Argument?	Can't Post

I'd like to do a search and replace on a bunch of filenames using regex values that are passed to the script from the command line argument which meand that the $1 or \1 will be in a string that I will be using in the regex.

For example:

Code

my $regexS = $ARGV[0]; 
my $regexR = $ARGV[1];

if $_ is something like 9999.txt

and I run the script

Code

scriptname '\d\d\d\d.txt' '00$0'

the line

Code

s/$regexS/$regexR/g;

turns $_; into

Code

00$0

instead of

Code

009999.txt

What can I to to make this work?

7stud
Enthusiast

Dec 3, 2009, 1:25 PM

Post #2 of 2 (1646 views)

Re: [ferulebezel] How Do I Use a regex back-reference in a Command Line Argument? [In reply to]	Can't Post

I don't know what you are trying to do with $0.

But I worked on your problem for a day using $1 and parentheses around part of the regex. The problem is that when perl reads a string from the command line, perl doesn't interpolate variables contained in the string.

Therefore, you need to force perl to eval the replacement string in a context where $1 is visible--then perl will interpolate $1 into the replacement string. I couldn't figure out how to get eval() to do that, so I asked for help. It turns out, you can do everything inside s///, but the solution is pretty ugly. I would do it like this instead:

Code

use strict; 
use warnings; 
use 5.010; 
 
my $str = "9999.txt"; 
my $pattern = shift; 
my $repl = shift; 
 
if ($str =~ $pattern) {#now $1 has a value 
 
    my $repl = eval qq{"$repl"}; #double double quotes are needed 
 
    $str =~ s/$pattern/$repl/; 
} 
 
say $str; 
 
--output:-- 
$perl 2perl.pl '(\d{4})' '00$1' 
009999.txt

=======

Code

$str =~ s/$pattern/qq{"$repl"}/ee; 
#or 
$str =~ s/$pattern/eval qq{"$repl"}/e;

(This post was edited by 7stud on Dec 3, 2009, 11:46 PM)

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