ajax访问json数据的属性值 - 编程论坛

问题：怎样取出json数据的某个属性值，比如我弹出框之弹出第一个对象的userName。

import
import
import java.util.ArrayList;
import java.util.List;

import javax.json.JsonArray;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

import com.alibaba.fastjson.JSON;
import com.alibaba.fastjson.JSONObject;
import com.pojo.Student;

public class TestJsonServlet extends HttpServlet {

    @Override
    protected void service(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {
        resp.setContentType("text/html;charset=utf-8");
        PrintWriter pw=resp.getWriter();

        List<Student> list=getList();
        String json=JSON.toJSONString(list);//将list转换成json

        pw.println(json);
        pw.close();
    }
    public List<Student> getList(){
        Student st1=new Student("张","男","666","324@);
        Student st2=new Student("王","女","123","324@);
        Student st3=new Student("何","女","456","324@);
        Student st4=new Student("贾","男","999","324@);
        List<Student> list=new ArrayList<>();
        list.add(st1);
        list.add(st2);
        list.add(st3);
        list.add(st4);
        return list;
    }

}

<html>
  <head>
    <base href="<%=basePath%>">
    <script type="text/javascript" src="js/jquery-2.0.0.min.js"></script>
  <script type="text/javascript">
  function test(){
  $.ajax({
  url:'testJson',
  type:'post',
  data:'json',
  success:function(data){
      alert(data.userName); //弹出的是defined，这个地方应该怎么写。


  }
  });
  }

  </script>
  </head>

  <body>
  <input type="button" value="注册" onclick="test();"/>
  </body>
</html>