javascript - 如何重写 XMLHttpRequest？

Question

我想重写“XMLHttpRequest”这个构造函数。

我只想在新建实例时发出警报，就像这样:(这不是我真正想做的，只是一个例子)

var ajax_request = new XMLHttpRequest(); //then alert "new a XMLHttpRequest instance!"

然后我试试这个:

var orig_XMLHttpRequest = window.XMLHttpRequest;
window.XMLHttpRequest = function() {
    alert("new a XMLHttpRequest instance!");
    orig_XMLHttpRequest.apply(this, arguments);
};

但是当我新建一个实例时，

我在 orig_XMLHttpRequest.apply(this, argument); 处收到此错误

TypeError: Constructor XMLHttpRequest requires 'new'

那么，我哪一步做错了？如何覆盖 XMLHttpRequest？还是不可能？

我也尝试这个:

var orig_XMLHttpRequest = window.XMLHttpRequest;
window.XMLHttpRequest = function() {
    alert("new a XMLHttpRequest instance!");
    new (Function.prototype.bind.apply(orig_XMLHttpRequest, arguments));
};

var ajax_request = new XMLHttpRequest();
ajax_request.open("POST", "./php/register.php", true);
ajax_request.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
ajax_request.onreadystatechange = function() {.....};
ajax_request.send();

但我仍然遇到错误:(

TypeError: ajax_request.open is not a function

Answer 1

有趣的问题。

我已经能够修改您的第二个代码示例以使其正常工作:

var orig_XMLHttpRequest = window.XMLHttpRequest;
window.XMLHttpRequest = function() {
  document.write("new a XMLHttpRequest instance!");
  return new orig_XMLHttpRequest();
};

var ajax_request = new XMLHttpRequest();
ajax_request.open("POST", "./php/register.php", true);
ajax_request.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
ajax_request.onreadystatechange = function() {
  if (ajax_request.readyState === XMLHttpRequest.DONE) {
    console.log("request returned status code: " + ajax_request.status);
  }
}
ajax_request.onerror = function(e) {
  console.log("request failed: " + e.target.status);
  }

ajax_request.send();