没有为类org.json.JSONObject找到序列化程序,也没有发现创建BeanSerializer的属性

时间:2014-09-26 14:19:01

标签: java json spring rest serialization

从Web服务获取JSON,Json Array作为响应

   [3]
   0:  {
   id: 2
  name: "a561137"
    password: "test"
  firstName: "abhishek"
   lastName: "ringsia"
    organization: "bbb"
      }-
    1:  {
      id: 3
  name: "a561023"
password: "hello"
     firstName: "hello"
   lastName: "hello"
     organization: "hello"
   }-
 2:  {
  id: 4
  name: "a541234"
  password: "hello"
 firstName: "hello"
  lastName: "hello"
  organization: "hello"
    }

在JsonArray中获取响应后,在读取Json数组的Json对象时出错:

List<User> list = new ArrayList<User>();
JSONArray jsonArr = new JSONArray(response);

for (int i = 0; i < jsonArr.length(); i++) {
    JSONObject jsonObj = jsonArr.getJSONObject(i);
    ObjectMapper mapper = new ObjectMapper();
    User usr=   mapper.convertValue(jsonObj, User.class);
    list.add(usr);
}
  

没有为类org.json.JSONObject找到序列化程序,也没有发现创建BeanSerializer的属性(为了避免异常,请禁用SerializationConfig.Feature.FAIL_ON_EMPTY_BEANS))

1 个答案:

答案 0 :(得分:2)

首先要接受它作为Json数组,然后在读取它的Object时必须使用Object Mapper.readValue,因为Json Object仍然在String中。

List<User> list = new ArrayList<User>();
JSONArray jsonArr = new JSONArray(response);

for (int i = 0; i < jsonArr.length(); i++) {
    JSONObject jsonObj = jsonArr.getJSONObject(i);
    ObjectMapper mapper = new ObjectMapper();
    User usr = mapper.readValue(jsonObj.toString(), User.class);      
    list.add(usr);
}