Retrieve the request URL from the XHR object

Question

Retrieve the request URL from the XHR object

Is it possible to retrieve the request URL from an XHR object? I noticed that the URL is visible in Firebug through the channel property, but unfortunately it cannot be queried using JavaScript.

javascript

Answer 1

Answer №1

Here is a clever trick that eliminates the need for a wrapper:

let xhrPrototype = XMLHttpRequest.prototype,
    originalOpen = xhrPrototype.open;

xhrPrototype.open = function (method, url) {
    this._url = url;
    return originalOpen.apply(this, arguments);
};

Example of how to use it:

let request = new XMLHttpRequest();
request.open('GET', '...', true);
alert(request._url); // This will show an alert box with '...'

Answer 2

Here is a clever trick that eliminates the need for a wrapper:

let xhrPrototype = XMLHttpRequest.prototype,
    originalOpen = xhrPrototype.open;

xhrPrototype.open = function (method, url) {
    this._url = url;
    return originalOpen.apply(this, arguments);
};

Example of how to use it:

let request = new XMLHttpRequest();
request.open('GET', '...', true);
alert(request._url); // This will show an alert box with '...'

Answer 3

Answer №2

For those utilizing jQuery, the "beforeSend" function in your AJAX request can be a powerful tool for modifying the jqXHR object. Here's an example:

$.ajax({
...
url: "http://another/url",
beforeSend: function(jqxhr, settings) { jqxhr.newURL = "http://another/url"; },
...
});

By adding jqXHR.newURL to the jqXHR object in this manner, you'll have access to it in any of the callback functions.

Answer 4

For those utilizing jQuery, the "beforeSend" function in your AJAX request can be a powerful tool for modifying the jqXHR object. Here's an example:

$.ajax({
...
url: "http://another/url",
beforeSend: function(jqxhr, settings) { jqxhr.newURL = "http://another/url"; },
...
});

By adding jqXHR.newURL to the jqXHR object in this manner, you'll have access to it in any of the callback functions.

Answer 5

Answer №3

It seems like I grasp the issue you're facing.

Creating a wrapper for your XHR objects to enable specific functionality shouldn't be too complex.

Here is a basic example to illustrate:

// It is assumed that GetXHR() generates a new XHR object, compatible with different browsers.

function HTTPGet(url, onStartCb, onCompleteCb)
{
  var xhr = GetXHR();
  // Create a custom xhr object containing the URL and the actual xhr object.
  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);
}

This code hasn't been extensively tested, but it should function properly. With some adjustments (such as error handling, parameter passing, etc.), you can likely transform this example into a fully operational solution.

Answer 6

It seems like I grasp the issue you're facing.

Creating a wrapper for your XHR objects to enable specific functionality shouldn't be too complex.

Here is a basic example to illustrate:

// It is assumed that GetXHR() generates a new XHR object, compatible with different browsers.

function HTTPGet(url, onStartCb, onCompleteCb)
{
  var xhr = GetXHR();
  // Create a custom xhr object containing the URL and the actual xhr object.
  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);
}

This code hasn't been extensively tested, but it should function properly. With some adjustments (such as error handling, parameter passing, etc.), you can likely transform this example into a fully operational solution.

Answer 7

Answer №4

As per the information on https://developer.mozilla.org/en/XmlHttpRequest, gaining access to the channel is possible with elevated privileges. However, it's important to note that the use of channel is not standard and may not be supported in all browsers. The W3C specifications found at http://www.w3.org/TR/XMLHttpRequest/ do not make any reference to channel or any method for "extracting the request URL". This leads me to believe that achieving this task consistently across different browsers may not be feasible.

Answer 8

As per the information on https://developer.mozilla.org/en/XmlHttpRequest, gaining access to the channel is possible with elevated privileges. However, it's important to note that the use of channel is not standard and may not be supported in all browsers. The W3C specifications found at http://www.w3.org/TR/XMLHttpRequest/ do not make any reference to channel or any method for "extracting the request URL". This leads me to believe that achieving this task consistently across different browsers may not be feasible.

Answer 9

Answer №5

Two useful properties that are worth considering include:

xhr2

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

This feature is not compatible with Internet Explorer.

fetch

If you utilize fetch, you can access Response.url - https://developer.mozilla.org/en-US/docs/Web/API/Response/url.

Please note that this functionality is not supported by Internet Explorer.

Answer 10

Two useful properties that are worth considering include:

xhr2

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

This feature is not compatible with Internet Explorer.

fetch

If you utilize fetch, you can access Response.url - https://developer.mozilla.org/en-US/docs/Web/API/Response/url.

Please note that this functionality is not supported by Internet Explorer.

Answer 11

Answer №6

This solution came in handy for me as I was working on a for loop that involved making URL requests.

Typically, when making a URL request and then calling the open method on the XMLHttpRequest object, you can set 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); // same as originalUrl
     }
};
xml.open("GET", originalUrl, true);
xml.url = originalUrl;
xml.send();

Answer 12

This solution came in handy for me as I was working on a for loop that involved making URL requests.

Typically, when making a URL request and then calling the open method on the XMLHttpRequest object, you can set 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); // same as originalUrl
     }
};
xml.open("GET", originalUrl, true);
xml.url = originalUrl;
xml.send();

Retrieve the request URL from the XHR object

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

Answer №6

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