Re: How to match a token not be quoted?

On Jan 28, 2008 9:09 PM, Zhao, Bingfeng <Bingfeng.Zhao@xxxxxx> wrote:
Hello,
I want to a cure regex that match following requirements: given $line =
'abc abc "abc abcc" abcc', I want to replace all instances of "abc" that
not in quotation with, say 'd', so I expect I get 'd d "abc abcc" dc'.
What should I write my regex? I try some and referred cook book also, no
solution. Thanks in advance.
snip

While this can be done, it is often easier to understand the following code

#!/usr/bin/perl

use strict;
use warnings;

my $string = 'abc abc "abc abcc" abcc';

my @tokens = split /(")/, $string;
my $in_string = 0;
for my $token (@tokens) {
$in_string = not $in_string if $token =~ /"/;
next if $in_string;
$token =~ s/abc/d/g;
}
$string = join "", @tokens;

print "$string\n";

If you really need to do it with a regex then I would consult perldoc
-q balanced and
http://perldoc.perl.org/perlfaq6.html#Can-I-use-Perl-regular-expressions-to-match-balanced-text?
.

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