javascript not passing variable

For some reason when I pass the "color" variable php cant pick it up- it just comes up blank. can someone please look at my code and see if im doing something wrong.. thanks guys

if ($id == 'theme')
      {
      echo '<form>';
      echo '<div class="color_option"><input type="radio" name="color" value="#DD597D" onClick="color_changer(this.value,\'center_info\',\'4001\');"/>Red';
      echo '<div class="color_option"><input type="radio" name="color" value="#0000FF" onClick="color_changer(this.value,\'center_info\',\'4001\');"/>Blue';
      echo '<div class="color_option"><input type="radio" name="color" value="#00FF00" onClick="color_changer(this.value,\'center_info\',\'4001\');"/>Lime';
      echo '</form>';
      }

function color_changer(color, divname, userid)
{
document.getElementById(divname).style.background=color;
connect_ajax()
xmlhttp.onreadystatechange=function()
      {
      if (xmlhttp.readyState==4 && xmlhttp.status==200)
            {
            document.getElementById("right_side").innerHTML=xmlhttp.responseText;
            /*theme(userid);*/
            return;
            }
      }
xmlhttp.open("GET","../functions/account_settings.php?id=changetheme&do="+userid+"&co="+color,true);
xmlhttp.send();
}



if ($id == 'changetheme')
      {
      
      $theme = $_GET['co'];
      include ('../config/config.php');
      mysql_pconnect ($host, $user, $pass) or die ('Unable To Connect to Database!');
      mysql_select_db('providers') or die ('Unable To Select Database');
      $qry = "UPDATE contractors SET theme ='$theme' WHERE id = '$userid' LIMIT 1";
      echo $qry;
      echo $theme;      
      //$result = mysql_query($qry) or die ('Error in query: $qry. ' . mysql_error());
      mysql_free_result($result);
      }