创建插件以检测可疑的 AJAX - create plugins to detect suspicious ajax

create plugins to detect suspicious ajax

本文关键字：AJAX 检测插件创建 | 更新日期: 2023-09-27

我想为Firefox制作浏览器扩展，以检测加载隐藏页面并重定向到新页面的网站的ajax代码，例如如果用户访问index.php其中ajax加载两个页面一个是隐藏页面.php并重定向到new.php。是否有任何其他解决方案可以在客户端检测此ajax。

if (xmlhttp.readyState==4 && xmlhttp.status==200)
                {
                    //document.getElementById("myDiv").innerHTML="";
                }
            }
            xmlhttp.open("GET","hidden.php",true);
            xmlhttp.send();
        }

.HTML

<a href="new.php" onclick="function();">click here</a>

document.addEventListener('DOMContentLoaded', function() {
    getCurrentTabUrl(function(url) {
        fetchData(url);
    });
});
function fetchData(url)
{
    var xhr = new XMLHttpRequest();
    xhr.open("GET", url, true);
    xhr.onreadystatechange=function()
    {
        if (xhr.readyState==4 && xhr.status==200)
        {
            var data = xhr.responseText;
            var  index = data.indexOf('XMLHttpRequest');
            if(index != -1){
                document.getElementById("status").innerHTML = "The page contains AJAX requests";
            }else{
                document.getElementById("status").innerHTML = "Page doesn't contains AJAX";
            }
            //document.getElementById("status").innerHTML = data;
        }
    }
    //xhr.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
    //xhr.setRequestHeader("Access-Control-Allow-Origin", "*");
    //xhr.setRequestHeader("Access-Control-Request-Method", "POST");
    xhr.send();
}
function getCurrentTabUrl(callback) {
    var queryInfo = {
    active: true,
    currentWindow: true
    };
    chrome.tabs.query(queryInfo, function(tabs) {
        var tab = tabs[0];
        var url = tab.url;
        console.assert(typeof url == 'string', 'tab.url should be a string');
        callback(url);
    });
}
just go through this code you will get the better help

您可以在用户脚本中修改XMLHttpRequest 的原型。

/* Save the old method somewhere, it may be useful if you want to allow some AJAX */
XMLHttpRequest.prototype._send = XMLHttpRequest.prototype.send;
/* Override the previous method to define yours */
XMLHttpRequest.prototype.send  = function () {
    /* Do stuff here */
    alert(1);
    /* Use this line if you want to continue the AJAX request */
    XMLHttpRequest.prototype._send.apply(this, arguments);
}