How to pass a variable in a URL using Perl?

Question

I'm trying to pass parameters in a URL. I don't know what's missing, I tried to see how the URL looks after executing this script.

my $request3 = HTTP::Request->new(GET => $sql_activation);

my $useragent = LWP::UserAgent->new();
$useragent->timeout(10);

my $response2 = $useragent->request($request3);
if ($response2->is_success) {

    my $res2 = $response2->content;

    if ($res =~ m/[#](.*):(.*)[#]/g) {
        my ($key, $username) = ($1, $2);
        print "[+] $username:$key \n\n";
    }
    else {
      print "[-] Error \n\n";
    }
}

my $link =
      "http://localhost/wordpress/wp-login.php?action=rp&key=" 
    . $key
    . "&login="
    . $username;

sub post_url {

    my ($link, $formref) = @_;

    my $ua = new LWP::UserAgent(timeout => 300);
    $ua->agent('perlproc/1.0');
    my $get = $ua->post($link, $formref);

    if ($get->is_success) {
        print "worked \n";
    }
    else {
        print "Failed \n";
    }
}

After executing the script the URL is like this

site/wordpress/wp-login.php?action=rp&key=&login=

What do you expect $key and $username to be? Can you show where they are set? — gpojd
– gpojd, Commented May 16, 2013 at 18:12
What's missing is any code to set the variables you are using to construct the URI. — Quentin
– Quentin, Commented May 16, 2013 at 18:13
(And mashing strings together to make URIs isn't a great idea, use the URI module.) — Quentin
– Quentin, Commented May 16, 2013 at 18:14
Adding use strict to your code would have given you a big clue to the problem. — Dave Cross
– Dave Cross, Commented May 17, 2013 at 9:47

Quentin · Accepted Answer · 2013-05-16 18:29:13Z

1

Perl has block level scope. You define $key and $username in the block following an if statement. They don't live beyond that.

You need to create them (with my) before that block.

# HERE
my ( $key, $username );
if ( $response2->is_success ) {
    my $res2 = $response2->content;
    if ( $res =~ m/[#](.*):(.*)[#]/g ) {
        # Don't say my again
        ( $key, $username ) = ( $1, $2 );
    }
    else { print "[-] Error \n\n"; }
}

answered May 16, 2013 at 18:29

Quentin

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1 Comment

gpojd Over a year ago

@gilardinho, I'd go a step further and add use strict; and use warnings; to the top of your script. It would have caught that bug for you.

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