Question

您好我是servlet中的新手，并尝试使用ajax将对象从javascript发送到servlet。 javascript代码如下所示：

$.ajax({
      url:'GetUserServlet',
      contentType: "application/json",
      data: JSON.stringify(response),
      type:'post',
      cache:false,
      success:function(data){
         //alert(data);
        $('#somediv').text("user info sent successfully"); 
      },
      error:function(){
          $('#somediv').text("some error occured"); 
      }
   }

）;

这里的回复是从facebook api收到的对象。它是：

reponse={ first_name: "Jhon", last_name: "Doe", id: "19862217575855" }

doPost的{{1}}方法定义为：

GetUserServlet

protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { Gson gson = new Gson(); User user = gson.fromJson(request.getParameter(response), User.class); System.out.println(user); }是包含User.class，first_name和last_name

的getter和setter的另一个类

但程序没有编译。我使用了很多语法更改但找不到正确的值。如何在servlet中获取id对象？

Answer 1

        // 1. get received JSON data from request
        BufferedReader br = new BufferedReader(new InputStreamReader(request.getInputStream()));
        String json = "";
        if(br != null){
            json = br.readLine();
        }

        // 2. initiate jackson mapper
        ObjectMapper mapper = new ObjectMapper();

        // 3. Convert received JSON to User
        User user = mapper.readValue(json, User.class);

        // 4. Set response type to JSON
        response.setContentType("application/json");            

       System.out.println("..user.." + user);

从ajax发送json数据并在servlet中解析

1 个答案: