how can I take a dynamic variable in value of JSON? - regex

Question

I need use the find and replace through regular expression like following
use strict;
no strict 'refs';
use warnings;
use JSON;
use Encode qw( encode decode encode_utf8 decode_utf8);
my $data =
{
"find_replace" => [
{ "find" => "(.+?)&",
"replace"=> "$1"
}
]
};
my $find_replace_arr = $data->{'find_replace'};
my $string = "http://www.website.com/test.html&code=236523";
my $find = $find_replace_arr->[0]->{find};
my $replace = $find_replace_arr->[0]->{replace};
$string =~ s/$find/$replace/isge;
print $string;
exit();
in this code, I only want to "http://www.website.com/test.html" from the string.
I am not able to get replace (key)'s value dynamically, which is $1.
You can Run the above code.
This code throw the Error Use of uninitialized value $1 in string

Answer 1

Some things to consider. First, the regex ([^&]+) may not give the desired result, as it is really going to capture and replace with the same capture.. resulting in the same output string (confusing I bet).
Next, the replace string "$1"has to be quoted again and e modifier has to be doubled.
So try this:
my $data =
{
"find_replace" => [
{ "find" => "^(.+?)&.*",
"replace"=> '"$1"'
}
]
};
my $find_replace_arr = $data->{'find_replace'};
my $string = "http://www.website.com/test.html&code=236523";
my $find = $find_replace_arr->[0]->{find};
my $replace = $find_replace_arr->[0]->{replace};
$string =~ s/$find/$replace/isgee;
print $string;
exit();
Notice the new regex, ^(.+?)&.* will match the entire string, but the capture (...) will be the result to replace.