【问题标题】:How to convert JSON String into Arraylist of Objects如何将 JSON 字符串转换为对象的 Arraylist
【发布时间】:2018-10-26 12:09:49
【问题描述】:

我有 JSON 字符串,从 HTTP 请求接收到:

     [  
       {  
      "id":15,
      "title":"1",
      "description":"desc",
      "user_id":152
     },
    {  
      "id":18,
      "title":"2",
      "description":"desc",
      "user_id":152
    },
    {  
      "id":19,
      "title":"tab3",
      "description":"zadanka",
      "user_id":152
   }
]

如何将其转换为对象的 ArrayList?

【问题讨论】:

    标签: java json arraylist


    【解决方案1】:

    你需要声明一个pojo

    class Data{
      String id;
      String title;
      String description;
      String userId;
      //Generate setter an getter
    }
    

    对 json 的迭代如下:

    JSONArray jsonArr = new JSONArray("[your JSON Stirng]");
        List<Data> dataList = new ArrayList<Data>();
        for (int i = 0; i < jsonArr.length(); i++) {
            JSONObject jsonObj = jsonArr.getJSONObject(i);
            Data data = new Data();
            data.setId(jsonObj.getString("id"));
            data.setTitle(jsonObj.getString("title"));
            data.setDescription(jsonObj.getString("description"));
            data.setUserId(jsonObj.getString("user_id"));
            dataList.add(data);
        }
    

    你还需要 json jar。可以从here下载

    【讨论】:

    【解决方案2】:

    使用 Gson

    Gson gson = new Gson();
            ArrayList<Object> listFromGson = gson.fromJson("json string",
                   new TypeToken<ArrayList<Object>>() {}.getType());
    

    使用杰克逊

        ObjectMapper mapper = new ObjectMapper();
        ArrayList<Object> listFromJackson = mapper.readValue("json string",
               new TypeReference<ArrayList<Object>>(){});
    

    如果你可以将 pojo 定义为

    public class Example {
    
        private Integer id;
        private String title;
        private String description;
        private Integer userId;
        // setters / getters
    }
    

    然后

    ArrayList<Example> listFromGson = gson.fromJson("json string",
                   new TypeToken<ArrayList<Example>>() {}.getType());
    
    ArrayList<Example> listFromJackson = mapper.readValue("json string",
                   new TypeReference<ArrayList<Example>>(){});
    

    此外,您应该更喜欢使用List 而不是ArrayList

    【讨论】:

      【解决方案3】:

      如果您使用的是 RestApi,则将注释 @RequestBody 与您的 pojo 类一起使用。

      @RequestMapping(value="/your api name", method=RequestMethod.POST)
      public ResponseData createUser(@RequestBody MyPojo myPojo){
      System.out.println("Creating User "+myPojo.toString());
      
      //Here you will able to access your request data from myPojo object
      }
      

      制作你的 pojo 类:

      public class MyPojo
      {
      private Data[] data;
      
      public Data[] getData ()
      {
          return data;
      }
      
      public void setData (Data[] data)
      {
          this.data = data;
      }
      
      @Override
      public String toString()
      {
          return "ClassPojo [data = "+data+"]";
      }
      }
      
      public class Data
      {
      private String id;
      
      private String title;
      
      private String description;
      
      private String user_id;
      
      public String getId ()
      {
          return id;
      }
      
      public void setId (String id)
      {
          this.id = id;
      }
      
      public String getTitle ()
      {
          return title;
      }
      
      public void setTitle (String title)
      {
          this.title = title;
      }
      
      public String getDescription ()
      {
          return description;
      }
      
      public void setDescription (String description)
      {
          this.description = description;
      }
      
      public String getUser_id ()
      {
          return user_id;
      }
      
      public void setUser_id (String user_id)
      {
          this.user_id = user_id;
      }
      
      @Override
      public String toString()
      {
          return "ClassPojo [id = "+id+", title = "+title+", description = "+description+", user_id = "+user_id+"]";
      }
      }
      

      【讨论】:

        【解决方案4】:

        除了@Sudhir,我推荐使用Gson

        Gson gson = new GsonBuilder().create();
         Data p = gson.fromJson(jsonString, Data.class);
        // Or to array.
         Data[] data = gson.fromJson(jsonString, Data[].class);
        

        【讨论】:

          猜你喜欢
          • 2018-11-16
          • 1970-01-01
          • 2018-01-06
          • 2015-09-11
          • 1970-01-01
          • 1970-01-01
          • 1970-01-01
          • 1970-01-01
          相关资源
          最近更新 更多