抽象类的自定义gson序列化

Question

也许我跑错了方向，但我有一个元素列表，我想读。

我有一个抽象基类，让我们称之为Person:

public abstract class Person {
    public int id;
    public String name;
}

现在我有两个可能的实现：

public class Hunter implements Person {
    public int skill;
    // and some more stuff
}

public class Zombie implements Person {
    public int uglyness;
    // and some more stuff
}

[
  {"id":1, "type":"zombie", "name":"Ugly Tom", "uglyness":42},
  {"id":2, "type":"hunter", "name":"Shoot in leg Joe", "skill":0}
]

return gson.getDelegateAdapter(this, resultType);

Answer 1

如何将此JSON读为列表？

一种可能是创建一个像工厂一样的自定义反序列化器。

第一步是定义这个反序列化器

class PersonJsonDeserializer implements JsonDeserializer<Person> {
    @Override
    public Person deserialize(JsonElement json, Type typeOfT, JsonDeserializationContext context) throws JsonParseException {
        String type = json.getAsJsonObject().get("type").getAsString();
        switch(type) {
            case "zombie":
                return context.deserialize(json, Zombie.class);
            case "hunter":
                return context.deserialize(json, Hunter.class);
            default:
                throw new IllegalArgumentException("Neither zombie or hunter");
        }
    }
}

public class GsonTest {
    public static void main(String[] args) {
        String json = "[\n" +
                "  {\"id\":1, \"type\":\"zombie\", \"name\":\"Ugly Tom\", \"uglyness\":42},\n" +
                "  {\"id\":2, \"type\":\"hunter\", \"name\":\"Shoot in leg Joe\", \"skill\":0}\n" +
                "]";

        Gson gson = new GsonBuilder().registerTypeAdapter(Person.class, new PersonJsonDeserializer()).create();

        Type type = new TypeToken<List<Person>>(){}.getType();

        List<Person> list = gson.fromJson(json, type);

        for(Person p : list) {
            System.out.println(p);
        }
    }
}

Zombie{id=1; name=Ugly Tom; uglyness=42}
Hunter{id=2; name=Shoot in leg Joe; skill=0}

class PersonJsonDeserializer implements JsonDeserializer<Person> {
    @Override
    public Person deserialize(JsonElement json, Type typeOfT, JsonDeserializationContext context) throws JsonParseException {
        String className = json.getAsJsonObject().get("type").getAsString();
        className = Character.toUpperCase(className.charAt(0)) + className.substring(1);
        try {
            return context.deserialize(json, Class.forName(className));
        } catch (ClassNotFoundException e) {
            throw new RuntimeException(e);
        }
    }
}