classes and array declaration problem

I made a normal slide out menu in javascript.
Now i am using php to make sure i can put it on all pages and all.

so I made a class called menu item.
<? php

class menuitem
{
var $name; //name
$submenu = array();//array of submenu items.
var $link; //link
var $currentposition;
var $id
var $linkinfo;

function menuitem($name, $submenu, $link,$prevposition,$id,$linkinfo)
{
$this->name = $name;
$this->link = $link;
for($i =0; $i<submenu; i++)
{
$this->submenu[i] = new submenuitem(""," ");
}
$this->currentposition = $prevposition + 40;
$this->id =$id;
$this->linkinfo;
}

function getposition()
{
return $this->currentpostion;
}

function ypslideout()
{
$size = 50*sizeof($this->submenu);
print("new ypSlideOutMenu(\"$this->name\"".", \"right\", 120,$this->currentposition , 150, $size);\n");
}

function link()
{
print("<a id=\"$this->id\""." href=\"$this->link\""." onmouseover=\"ypSlideOutMenu.showMenu(\'$this->name\')\""." onmouseout=\"ypSlideOutMenu.hideMenu(\'$this->name\')\">"."$linkinfo"."</a><br>\n");
print("<br>");
}

function submenulink($array, $name, $link)
{
$this->submenu[$array]->setlink($name, $link);

}

function createsubmenu()
{
if(sizeof($this->submenu)!=0)
{
$divid1 = $name."container";
$divid2 = $name."content";
print("<div id = \"$divid1\">\n");
print(" <div id=\"$divid2\"." class=\"menu\">\n");
print(" <div class=\"options\">\n");
for(int i = 0;
print(" <a href="$this->submenu[i]->getlink()"".">$this->submenu[i]->getname()</a>\n");
print(" </div>\n");
print(" </div>\n)";
}
}?>

now i am getting a parse error on line 3. I wonder if something is wont with my array declaration.

Put your code in PHP tags so folks can actually read it while you're at it...

You're mixing your javascript and php code. Take a closer look at line three. There's only two 'words'...php doesn't have a var statement.

Actually - it does, he/she is intializing/defineing the attributes, thats how to do it in php.

The problem is initializing the att to array() - initialize it to nothing then asign it in the constructor...

can you please rexplain this to me?
I am not understanding.

The explaination is to move your question to the PHP forum, and everything will work again

Crap. Don't post on your birthday when you get back from the bar...