Question

I have a flash code, in which i am loading images and opening urls through xml file, now it is opening in a different window(pop up window) but i want it to open in the same window.
Please have a look at my code and suggest me.

var xmlReq:URLRequest = new URLRequest("XML/ImagesData01.xml"); 
var xmlLoader:URLLoader = new URLLoader();

xmlLoader.addEventListener(Event.COMPLETE, xmlLoaded); 
xmlLoader.load(xmlReq);

var myXML:XML;
var mySprite:Sprite;
var headLoader:Loader;


function imagepress(e:MouseEvent):void  
{
navigateToURL(new URLRequest(String(myXML.data[Number(e.currentTarget.name)].@url)),"_self");

}


function xmlLoaded(e:Event):void{
myXML = new XML(e.target.data);
for (var i:int = 0; i < myXML.children().length(); i++)
{
headLoader = new Loader();
mySprite = new Sprite();
headLoader.load(new URLRequest(String(myXML.data[i].@image)));
mySprite.addChild(headLoader);
mySprite.name = String(i);
mySprite.addEventListener(MouseEvent.CLICK, imagepress)
addChild(mySprite);
}


}

Answer 1

it’s not:

navigateToURL(new URLRequest(String(myXML.data[Number(e.currentTarget.name)].@url,"_self"))); 

rather it’s

navigateToURL(new URLRequest(String(myXML.data[Number(e.currentTarget.name)].@url)) ,"_self"); 

you’ve been sending 2 arguments to URLRequest class instead of navigateToURL function…
my code should work.