Return ID of all iframes in a page

Question

Because of the widget format I\'m working with I have a page which has multiple iframes embedded within iframes. I won\'t paste the code as it\'s vast and unwieldy but it is

Answer 1

From the most embedded iframe:

var ids = (function up( context, ids ){ 
   ids = ids || []; 
   // proceed if we're not at the top window yet
   if( context !== window.top){
      // point context to parent window (or top if no parent).
      context = context.parent || window.top; 
      // get the id of the first iframe in parent window; 
      // this will break if there are sibling iframes
      ids.push(context.document.getElementsByTagName('iframe')[0].id); 

      // recursive call to traverse parents          
      return up(context, ids); 
   } else {
      // otherwise return the list of ids
      return ids; 
   }
   }(this)); // 'this' is the initial context - current window

   console.log( ids );

Answer 2

I do not believe you can reference the iframe's children directly. You will need to recursively search each iframe using it's .contents() call.

As far as I know, this will only work as long as the same-origin policy is not violated (i.e. the iframes must point to the same domain).

Answer 3

EDITED

This returns an iframe element which has the wanted id, and null, if the wanted id is not found.

function searchIds(wantedId) {
    var idArray = [], n,
        search = function (iframes) {
            var n;
            for (n = 0; n < iframes.length; n++) {
                if (iframes[n].frames.length > 0) {
                    search(iframes[n].frames);
                }
                idArray.push(iframes[n].frameElement.id);
                idArray.push(iframes[n].frameElement);
            }
        };
    search(window.top.frames);
    for (n = 0; n < idArray.length; n += 2) {
        if (idArray[n] === wantedId) {
            return idArray[n + 1];
        }
    }
    return null;
}

Notice, that searchIds() can't be run before onload of the main window has been fired.