How to get the URL of a xmlhttp request (AJAX)

问题:

On w3schools.com(url) there is an example of how to do an AJAX call with plain Javascript. If you look at the example you will see the call is triggered by a button:

<button type="button" onclick="loadXMLDoc()">Change Content</button>

This is the function:

function loadXMLDoc()
{
var xmlhttp;
if (window.XMLHttpRequest)
  {// code for IE7+, Firefox, Chrome, Opera, Safari
  xmlhttp=new XMLHttpRequest();
  }
else
  {// code for IE6, IE5
  xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
  }
xmlhttp.onreadystatechange=function()
  {
  if (xmlhttp.readyState==4 && xmlhttp.status==200)
    {
    document.getElementById("myDiv").innerHTML=xmlhttp.responseText;
    }
  }
xmlhttp.open("GET","ajax_info.txt",true);
xmlhttp.send();
}

What I would like to do is get the URL of the outgoing AJAX call which is ajax_info.txt(url):

xmlhttp.open("GET","ajax_info.txt",true);

Im trying to put that URL in to an alert, so I tried calling the headers of the response using getAllResponseHeaders() hoping that it will give me the Host like so:

if (xmlhttp.readyState==4 && xmlhttp.status==200)
    {
    document.getElementById("myDiv").innerHTML=xmlhttp.responseText;
    alert(xmlhttp.getAllResponseHeaders());

It does give me all the headers but not the Host. So my next move was trying to set the Host myself using setRequestHeader() but then I realized the Header needs a Value which I had to send myself, so this will not work. What else can I try to get/fetch the outgoing AJAX URL in the alert?

Please note the code is just an example and I know that changing headers(in this case) is prohibited because of Access-Control-Allow-Origin.

回答1:

I'm not sure how much access you have to the code but you can over-ride XMLHttpRequest.open and hook the url there.

XMLHttpRequest.prototype.open = (function(open) {
  return function(method,url,async) {
      console.log('the outgoing url is ',url);
      open.apply(this,arguments);
    };
})(XMLHttpRequest.prototype.open);

Here is a FIDDLE.