I am trying to use ajax and getting this error message and do not understand Why

[ragtek]

hi

i think you'll find all the answers on the yui page http://developer.yahoo.com/yui/connection/

here's a example how i'm using it:

var invites = {
		init:function()
		{
			YAHOO.util.Event.removeListener("ajaxinvites", "click", invites.init);
			YAHOO.util.Connect.asyncRequest("POST", "ajax.php?do=showinviteform", {
				success: invites.initform,
			    failure: vBulletin_AJAX_Error_Handler,
			    timeout: vB_Default_Timeout
			},SESSIONURL + "securitytoken=" + SECURITYTOKEN + "&do=showinviteform"); 
			
		},
		initform:function(a)
		{
			b = a.responseXML.getElementsByTagName("template")[0].firstChild.nodeValue;
			t = YAHOO.util.Dom.get('ajaxinvites').innerHTML;
			YAHOO.util.Dom.get('ajaxinvites').innerHTML = b;
			YAHOO.util.Event.on('sendinvite', 'click', invites.send, this, true);
		},
.....}

second example:

vBTwitter.prototype.senddirect = function(e)
{
	YAHOO.util.Event.stopEvent(e);
	YAHOO.util.Dom.get("progress_tweet").style.display='block';
	YAHOO.util.Connect.asyncRequest("POST", "tweet.php?do=tweet", {
		success: this.sended,
	    failure: vBulletin_AJAX_Error_Handler,
	    timeout: vB_Default_Timeout
	},SESSIONURL + "securitytoken=" + SECURITYTOKEN + "&do=tweet&tweet=" + PHP.urlencode(YAHOO.util.Dom.get('twitter_tweet').value) );  // TODO need to check which function to use here to code msg
}

vBTwitter.prototype.sended = function(a)
{
		var error = a.responseXML.getElementsByTagName("error")[0];
		if (error)
		{
			YAHOO.util.Dom.get("progress_tweet").style.display='none';
			alert(a.responseXML.getElementsByTagName("error")[0].firstChild.nodeValue);
		}
		else
		{
			YAHOO.util.Dom.get("progress_tweet").style.display='none';
			YAHOO.util.Dom.get('twitter_tweet').value = '';
			vBmenu.hide();
		}
}

Maybe its not the best way to use it, but thats the way i'm using it and its working