如何使用Ajax(XMLHttpRequest)发送带参数的请求，以及如何在Servlet中获取请求中的参数

来源：互联网发布：海南前景知乎编辑：程序博客网时间：2024/05/16 05:29

其实这样设计的初衷是为了在客户端发出请求前，对要发送的数据进行预处理。比如用户密码的加密操作等等。

var xmlhttp;//设置全局变量function login() {        //这里为了简化代码，没有附上密码加密的代码var params = "username=" + document.getElementById("username").value+ "&password=" + document.getElementById("password").value;sendRequest("POST","login",true,params,function() {if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {document.getElementById("message").innerHTML = xmlhttp.responseText;}});}function sendRequest(method, url, isAsyns, params, action) {if (window.XMLHttpRequest) {// code for IE7+, Firefox, Chrome, Opera, Safarixmlhttp = new XMLHttpRequest();} else {// code for IE6, IE5xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");}xmlhttp.open(method, url, isAsyns);xmlhttp.setRequestHeader("Content-type","application/x-www-form-urlencoded");//这行代码很关键，用来把字符串类型的参数序列化成Form Dataxmlhttp.send(params);xmlhttp.onreadystatechange = action;}

public void doPost(HttpServletRequest request, HttpServletResponse response) {String username = request.getParameter("username");String password = request.getParameter("password");......}