Question

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

Answer 1

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, Safari xmlhttp = new XMLHttpRequest(); } else {// code for IE6, IE5 xmlhttp = 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 Data xmlhttp.send(params); xmlhttp.onreadystatechange = action; }public void doPost(HttpServletRequest request, HttpServletResponse response) { String username = request.getParameter("username"); String password = request.getParameter("password"); ...... }