инклюд PHP кода в шаблон - Joomla 1.0: Общие вопросы

<div><?php include('menu.php')?></div>

<div id="topnavi">
<ul id="mainlevel-nav">
<?php
$item_id = mysql_escape_string( $_GET['Itemid'] );
$qry = "SELECT id, name, link FROM #__menu WHERE menutype='mainmenu' and parent='0' AND access<='$gid' AND sublevel='0' AND published='1' ORDER BY ordering";
$database->setQuery($qry);
$rows = $database->loadObjectList();
foreach($rows as $row) {
echo "<li><a href='$row->link&Itemid=$row->id' ".( $row->id == $item_id ? "class='current'" : "" )." ><span>$row->name</span></a></li>";
}
?>
</ul>
</div>