在GSON中为类的每个实例添加属性

Question

class Human {
   String name;
}

class Student extends Human {
   String college;
}

class Worker extends Human {
   String workPlace;
}

假设我想使用GSON序列化它。

是否可以为每个序列化"type" : "student"实例添加一对Student（就像type是该类的字段一样）？同样，为每个"type" : "worker"实例添加Worker

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相关问题涉及此类JSONS的反序列化： Deserialize recursive polymorphic class in GSON

Answer 1

你可以通过gson自定义JsonSerializer这样做

public class HumanSerializer implements JsonSerializer<Human> {

 @Override
public JsonElement serialize(final Human human, final Type type, final JsonSerializationContext context) {
             final JsonObject json = new JsonObject();
             if(human instanceof Human)
                 json.addProperty("type", "Human");
             if(human instanceof Worker)
                 json.addProperty("type", "Worker");
             if(human instanceof Student)
                 json.addProperty("type", "Student");

         json.addProperty("name", human.getName());
        return json;
    }
}

最后你必须注册你的课程然后序列化

final GsonBuilder  gsonBuilder = new GsonBuilder();
gsonBuilder.registerTypeAdapter(Human.class,new  HumanSerializer());
gsonBuilder.registerTypeAdapter(Worker.class,new  HumanSerializer());
gsonBuilder.registerTypeAdapter(Student.class,new  HumanSerializer());
final Gson gson = gsonBuilder.create();

输出

 gson.toJson(new Worker("adam", "workplace"));
 gson.toJson(new Human("Jhon"));

{"type":"Worker","name":"adam"}
{"type":"Human","name":"Jhon"}