jQuery的Ajax实现异步传输List、Map.doc

jQuery的Ajax实现异步传输List、Map 由于项目需要通过ajax得到后台的List对象里面的值，由于没有接触过json所以前期我只能先传字符串再在js里截取子串实现。闲暇时间摸索了一下关于json接受List对象的方法，步骤如下： 1.首先下载json所以依赖的包， Json-lib 必威体育精装版版 json-lib-2.3-jdk15.jar，其官方网站是：/可以直接download，细心的你会发现其网站页面里面有提示它还需要其他包： Json-lib requires (at least) the following dependencies in your classpath: jakarta commons-lang 2.4 jakarta commons-beanutils 1.7.0 jakarta commons-collections 3.2 jakarta commons-logging 1.1.1 ezmorph 1.0.6?? 其中一般项目中都会有前4个，所以只需要下载ezmorph 1.0.6? 就可以了 下载地址分别是： /lang/ /beanutils/ /collections/ /logging/ / 2.首先先见一个小web项目 先建一个User类： package com.json; public class User { ??? String username; ??? String password; public String getUsername() { ? return username; } public void setUsername(String username) { ? this.username = username; } public String getPassword() { ? return password; } public void setPassword(String password) { ? this.password = password; } ??? } 再建一个servlet package com.json; import java.io.IOException; import java.io.PrintWriter; import java.util.ArrayList; import java.util.List; import javax.servlet.ServletException; import javax.servlet.http.HttpServlet; import javax.servlet.http.HttpServletRequest; import javax.servlet.http.HttpServletResponse; import net.sf.json.JSONArray; public class TestJson extends HttpServlet { public void doGet(HttpServletRequest request, HttpServletResponse response) ?? throws ServletException, IOException { ?????????? doPost(request,response); } public void doPost(HttpServletRequest request, HttpServletResponse response) ?? throws ServletException, IOException { ? response.setContentType(text/html); String str= request.getParameter(name);//得到ajax传递过来的paramater ? System.out.print(str); ? PrintWriter out = response.getWriter(); ? List list = new ArrayList();//传递List ????????????????Map m=new HashMap();//传递Map??? ? User u1=new User(); ? u1.setUsername(zah); ? u1.setPassword(123); ? User u2=new User(); ? u2.setUsername(ztf); ? u2.set