Запустите этот PHP-файл через Cron Job

Вот скрипт:

<?php
//Create back files?
define('CREATE_BACKUPS', FALSE);

if (!is_dir($argv[1]))
{
   echo "You must enter a valid path such as /home/apresv/public_html or apresv/public_html for this script to function.\n";
   exit;
}

//Search the path for all php files, opening each one, and checking to see if it's infected

//First, get an array list of all valid .php files.


$files = listdir($argv[1]);
foreach ($files as $filename)
{
   //We only need to check php files, so we add that here
   if (file_extension($filename) == 'php')
   {
      //This is a php file so lets check it to see if it's infected.
      $contents = file_get_contents($filename);
      $backup = $contents;

      //There will always be 2 opening tags in an infected file and since the hack is always at the top, it's easiest to test for this right away.
      $test = between('<?php', '<?php', $contents);

      //This particular hack likes to use toolbarqueries so we test to see if our chunk is an infected chunk.  If your website uses this url somehow, then add extra if statements as necessary.
      if (after('toolbarqueries', $test))
      {
         //This chunk is infected.  So lets replace it and resave the file.
         $contents = str_replace('<?php'.$test.'<?php', '<?php', $contents);

         //Now save it! Woohoo!
         file_put_contents($filename, $contents);
         if (CREATE_BACKUPS)
         {
            file_put_contents($filename.'.orig', $backup);
         }

         echo "$filename has been cleaned.\n";
      }
   }
}

function after ($this, $inthat)
    {
        if (!is_bool(strpos($inthat, $this)))
        return substr($inthat, strpos($inthat,$this)+strlen($this));
    };

    function after_last ($this, $inthat)
    {
        if (!is_bool(strrevpos($inthat, $this)))
        return substr($inthat, strrevpos($inthat, $this)+strlen($this));
    };

    function before ($this, $inthat)
    {
        return substr($inthat, 0, strpos($inthat, $this));
    };

    function before_last ($this, $inthat)
    {
        return substr($inthat, 0, strrevpos($inthat, $this));
    };

    function between ($this, $that, $inthat)
    {
     return before($that, after($this, $inthat));
    };

    function between_last ($this, $that, $inthat)
    {
     return after_last($this, before_last($that, $inthat));
    };

    // USES
    function strrevpos($instr, $needle)
    {
        $rev_pos = strpos (strrev($instr), strrev($needle));
        if ($rev_pos===false) return false;
        else return strlen($instr) - $rev_pos - strlen($needle);
    };

    function listdir($dir='.') {
    if (!is_dir($dir)) {
        return false;
    }

    $files = array();
    listdiraux($dir, $files);

    return $files;
}

function listdiraux($dir, &$files) {
    $handle = opendir($dir);
    while (($file = readdir($handle)) !== false) {
        if ($file == '.' || $file == '..') {
            continue;
        }
        $filepath = $dir == '.' ? $file : $dir . '/' . $file;
        if (is_link($filepath))
            continue;
        if (is_file($filepath))
            $files[] = $filepath;
        else if (is_dir($filepath))
            listdiraux($filepath, $files);
    }
    closedir($handle);
}

function file_extension($filename)
{
   $info = pathinfo($filename);
   return $info['extension'];
} 
?>

Когда я пытаюсь запустить его через cron, я получаю вывод "Вы должны ввести правильный путь, такой как /home/apresv/public_html или apresv/public_html, чтобы этот скрипт работал".

Что мне нужно сделать, чтобы это запускалось через CRON JOB?

Спасибо!