window.XMLHttpRequest.prototype.open 可以监听 XMLHttpRequest 。但不能监听fetch请求。
<!DOCTYPE html><html lang="en"><head> ???<meta charset="UTF-8"> ???<title>Document</title> ???<meta name="viewport" content="width=device-width, initial-scale=1"> ???<!-- jquery --> ???<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script></head><body></body><script>var _open = window.XMLHttpRequest.prototype.openvar _send = window.XMLHttpRequest.prototype.send// https://github.com/Tencent/vConsole/blob/dev/src/network/network.js#L271window.XMLHttpRequest.prototype.open = function() { ???console.log(20181117200649, arguments) ???return _open.apply(this, arguments)}// fetch("http://localhost/fuck.php").then(response => {// ????// console.log(20181117201438, response.json())// ????return response.json()// }).then(data=>{// ????console.log(20181117201450, data)// })$.ajax({ ???url: "http://localhost/fuck.php", ???success: function (data) { ???????console.log(20181117133228, data); ???}})</script></html>温故而知新 监听 XMLHttpRequest 发起请求
原文地址：https://www.cnblogs.com/CyLee/p/9977143.html