ajax如何向action发送数据的

首先定义一个javascript函数，用来验证注册时用户名是否已经存在

function validateName(){    //validateName值得是structs.xml中的名字是validateName那个action，在spring配置文件中注入var url = 'validateName!execute';var json = document.getElementById("name").value;//获取用户名        var jsonName = {name:json}//创建json对象var name = JSON.stringify(jsonName); //将json对象转换成json格式的字符串            $.post(url, {json: name}, function(json){        $("#tip").text(json.msg); }, "json");}

看看具体的ValidateAction实现

package org.arthur.lr.system.action;import java.io.IOException;import java.io.PrintWriter;import javax.servlet.http.HttpServletResponse;import org.apache.struts2.ServletActionContext;import org.arthur.lr.system.dao.UserManager;import org.json.JSONObject;import com.googlecode.jsonplugin.annotations.JSON;import com.opensymphony.xwork2.ActionSupport;public class ValidateAction extends ActionSupport {/** *  */private static final long serialVersionUID = 1L;//这个属性已经在spring配置文件中注入，千万不要在这里写UserManger mgr = new UserMangerImpl();private UserManager mgr;private String json;@JSONpublic String getJson() {return json;}public void setJson(String json) {this.json = json;}public UserManager getMgr() {return mgr;}public void setMgr(UserManager mgr) {this.mgr = mgr;System.out.println("启动tomcat时初始化");}@Overridepublic String execute() throws Exception {JSONObject jsonObj = new JSONObject(json);    //将JSON格式的字符串构造成JSON对象        String name = jsonObj.getString("name"); //获取JSON对象中的name属性的值        if (mgr.validateName(name)) {  json =  "{msg:'you can use this name to regist'}";System.out.println("有户名有效");} else { json =  "{msg:'the name is already exits'}";System.out.println("用户名已经存在");}this.sendMsg(json);//发送数据到客户端，供回调函数调用return SUCCESS;}public void sendMsg(String msg) throws IOException{System.out.println("发送json");HttpServletResponse res = ServletActionContext.getResponse();PrintWriter out = res.getWriter();                   //out.println(msg);这样写也行out.write(msg);if(null != out){ out.close();//如果不关闭的话，ajax的回调函数不会调用}       }}