java - 解析具有后代元素的 JSON 列表

Question

有一个函数返回 JSON 格式的元素列表。从包含后代类型元素的结果中读取内容时出现问题。

例如，考虑以下定义:

public class Fruits {
    public String description;
    public List<Fruit> fruits = new ArrayList<Fruit>();
}

public class Fruit {
    public String name;
    public int pieces;
}

public class Banana extends Fruit {
    public String origin;
}


@ApplicationScoped
@Provider
@Produces("application/json")
@Path("/fruit")
public class MyRestService {
    @GET
    @Path("/fruits")
    public Fruits getFruits() {
        Fruits myFruits = new Fruits();
        myFruits.description = "My fruits";

        Fruit myApple = new Fruit();
        myApple.name = "apple";
        myApple.pieces = 8;
        myFruits.fruits.add(myApple);

        Banana myBanana = new Banana();
        myBanana.name = "banana";
        myBanana.pieces = 5;
        myBanana.origin = "Ecuador";
        myFruits.fruits.add(myBanana);

        return myFruits;
    }
}

生成的 JSON 如下:

{  
   "description":"My fruits",
   "fruits":[  
      {  
         "name":"apple",
         "pieces":8
      },
      {  
         "name":"banana",
         "pieces":5,
         "origin":"Ecuador"
      }
   ]
}

假设我们在 content 变量中有这个结果。我想按如下方式解析它:

ObjectMapper objectMapper = new ObjectMapper();
Fruits fruits = objectMapper.readValue(content, Fruits.class);

抛出以下异常:

Exception in thread "main" org.codehaus.jackson.map.exc.UnrecognizedPropertyException: Unrecognized field "origin" (Class myresttest.Fruit), not marked as ignorable
 at [Source: java.io.StringReader@38653ad5; line: 1, column: 104] (through reference chain: myresttest.Fruits["fruits"]->myresttest.Fruit["origin"])

是否有可能在不接触服务器代码的情况下解决这个问题？

Answer 1

我建议你使用json-simple-1.1.1

您可以像这样获取 JSonObject :

JSONObjec requestobj = (JSONObject) JSonValue.parse(request) ;

Note: request value should be a String representation of the JSON .

然后将 requestobj 字段解析为一个新的 Fruits 实例...... 例如:

Fruits myFruits = new Fruits();

myFruits.description = (String) requestobj.getString("description");