error using XmlHttpRequest in Firefox

[satishr23]

Hi, I am trying to develop a RSS reader using Ajax.
I use the following Get statement.
AjaxObject.open ("GET",URL,true )
where URL is the location of my feed.
Now when I give URL as a file in my localhost directory, it works fine in both IE and Firefox.
But when give URL as a remote feed,it doesnt work in Firefox. It gives me an 'undefined' error.
I read somewhere that it could be because Firefox blocks cross-domain operations so I included the folowing statement in my Js file

try
{
netscape.securi ty.PrivilegeMan ager.enablePriv ilege("Universa lBrowserRead");
}
catch(e)
{
alert("Permissi on UniversalBrowse rRead denied.");
}

But it still goes into the catch mode...

Any idea on how to solve this problem?
Help would be greatly appreciated.
Thanks

[dmjpro]

can u send me ur js code where u make the request.
then i think it ll be better to solve ur problem.

kind regards.
dmjpro.

[satishr23]

Hi,
Thanks for your reply..
Here is my code..

url="http://newsrss.bbc.co.uk/rss/newsonline_uk_edition/front_page/rss.xml";
function getRSS()
    { 
	if (window.ActiveXObject)
		{
		ajaxobj = new ActiveXObject("Microsoft.XMLHTTP");
		alert("microsoft xmlhttp");
		}
	else if (window.XMLHttpRequest)
		{
		ajaxobj = new XMLHttpRequest();
		if (typeof netscape != 'undefined' && typeof netscape.security !=             'undefined') 
                {
                 try {
          netscape.security.PrivilegeManager.enablePrivilege('UniversalBrowserRead');
                     }
                catch (e) {
                               alert('Error ' + e.message + ' occurred.');
                             }
		 }
         
	       alert("xmlhttprequest");
	     }
	else
		alert("not supported");
		
	try
	{
	ajaxobj.open("GET",url+"?sid="+Math.random(),true);
	}
	catch(error)
	{
	 alert("catch"+ error.description);   //script doesnt go beyond this in Firefox
	 return false;
	 }
       ajaxobj.onreadystatechange = function() {
		if (ajaxobj.readyState == 4)
		{
			if (ajaxobj.status == 200)
			{
				if (ajaxobj.responseText != null)
				{
				processRSS(ajaxobj.responseXML);
				}
				else
				{
				 alert("Failed to receive RSS file from the server - file not found.");
				return false;
				}
			}
		 else
			alert("Error code " + ajaxobj.status + " received: " + ajaxobj.statusText);
		 }
	}

	ajaxobj.send(null);
    }

[acoder]

To get around the security (same domain), use JSON, e.g. link.