Get request url from xhr object

I hope I'm understanding your problem correctly.

It should be fairly simple to wrap your XHR objects with your own object to allow for this kind of functionality.

Below is an overly simplified example:

// Assumption: GetXHR() returns a new XHR object, cross browser.

function HTTPGet(url, onStartCb, onCompleteCb)
{
  var xhr = GetXHR();
  // Construct your own xhr object that contains the url and the xhr object itself.
  var myXhr = { xhr: xhr, url: url };

  xhr.onreadystatechange = function()
  {
     if (xhr.readyState == 4 && xhr.status == 200)
     {
        onCompleteCb(myXhr);
     }
  };

  xhr.open("GET", url);
  onStartCb(myXhr);
  xhr.send(null);
}

I haven't tested this extensively, but it should work and with some modifications (error handling, passing parameters, etc) you should probably be able to turn this example into a fully functional solution.

According to https://developer.mozilla.org/en/XmlHttpRequest, you can indeed get channel if you have elevated privileges; however, channel is non-standard (might not work in other browsers), and indeed the W3C specs at http://www.w3.org/TR/XMLHttpRequest/ do not mention channel nor any other way to "extract the request URL". I suspect, therefore, that there is no way to do so reasonably across browsers.

A couple of related properties that could be useful are:

xhr2

XMLHttpRequest.responseURL - https://developer.mozilla.org/en-US/docs/Web/API/XMLHttpRequest/responseURL.

This is not supported by IE.

fetch

If you are using fetch you can use Response.url - https://developer.mozilla.org/en-US/docs/Web/API/Response/url.

This is not supported by IE.

I also needed this because I was executing a for loop where I did URL requests.

Since usually you have the URL request before you call the open method on the XMLHttpRequest object, you can assign a random property on the xml object (in my case, the property url):

var xml = new XMLHttpRequest();
xml.onreadystatechange = function() {
    if ( xml.readyState == 4) {
        console.log(xml.url); // equal to originalUrl
    }
};
xml.open("GET", originalUrl, true);
xml.url = originalUrl;
xml.send();