0

I currently searching for a code/script that will allow it to check if a specific port is open, let's say 123. Found a public script on a ftp but I don't know how and what to modify in it to suit my needs. (I think this is a evil code and I want to use it as an example).

cat $1 | while read LN; do
SRV=`echo $LN | awk '{print $1}'`
php connect.php $SRV 123
done

And the file connect.php

<?php

    function S_Server($_server, $_port, $_user, $_pass) {
    if ( "$_port" == "123" ) {
            print "[+]Found $_server\r\n";
            if(!($_OutFile = fopen("iplist", "a"))) ExitF ("Cannot open the log file");
            fputs($_OutFile, "$_server $_user $_pass\n");
            fclose($_OutFile);
    }

    }

    function ExitF($errmsg) {
    print "[-]" . $errmsg . "\r\n";
    exit(0);
    }

    function CrackSMTP($server, $port, $user, $pass) {
    $socket = fsockopen($server, $port, $errno, $errstr, 2);

    if (!$socket) ExitF ("SOCKET ERROR!");

    fclose($socket);

    S_Server ($server, $port, $user, $pass);

    exit(0);
    }

    if ($argv[4]) $_PASS = $argv[4];
    if (!($_PASS)) $_PASS = "";

    if (!($argv[3])) {
    ExitF ("Usage: $argv[0] <hostname> <port> <user> [password]");
    }

    else {
    CrackSMTP($argv[1], $argv[2], $argv[3], $argv[4]);
    }

    exit(0);
    ?>
Peter O.
  • 32,158
  • 14
  • 82
  • 96
Cristian-Dumitru Popescu
  • 17
  • 8

1 Answers1

1

Try this:

fsockopen(hostname,port)

file.php:

<?php

if(fsockopen($argv[1], $argv[2]))
{
  print "I can see port $argv[2] from host $argv[1]";
}

Example: php file.php google.com 80

Output:

I can see port 443 from host google.com
Cyrus
  • 84,225
  • 14
  • 89
  • 153