如何使用接口序列化类？

json interface java gson

2022-09-01 04:49:34

我从来没有做过太多的序列化，但我试图使用Google的gson将Java对象序列化为文件。以下是我的问题的一个示例：

public interface Animal {
    public String getName();
}


 public class Cat implements Animal {

    private String mName = "Cat";
    private String mHabbit = "Playing with yarn";

    public String getName() {
        return mName;
    }

    public void setName(String pName) {
        mName = pName;
    }

    public String getHabbit() {
        return mHabbit;
    }

    public void setHabbit(String pHabbit) {
        mHabbit = pHabbit;
    }

}

public class Exhibit {

    private String mDescription;
    private Animal mAnimal;

    public Exhibit() {
        mDescription = "This is a public exhibit.";
    }

    public String getDescription() {
        return mDescription;
    }

    public void setDescription(String pDescription) {
        mDescription = pDescription;
    }

    public Animal getAnimal() {
        return mAnimal;
    }

    public void setAnimal(Animal pAnimal) {
        mAnimal = pAnimal;
    }

}

public class GsonTest {

public static void main(String[] argv) {
    Exhibit exhibit = new Exhibit();
    exhibit.setAnimal(new Cat());
    Gson gson = new Gson();
    String jsonString = gson.toJson(exhibit);
    System.out.println(jsonString);
    Exhibit deserializedExhibit = gson.fromJson(jsonString, Exhibit.class);
    System.out.println(deserializedExhibit);
}
}

因此，这可以很好地序列化 - 但可以理解的是，删除了动物上的类型信息：

{"mDescription":"This is a public exhibit.","mAnimal":{"mName":"Cat","mHabbit":"Playing with yarn"}}

但是，这会导致反序列化的实际问题：

Exception in thread "main" java.lang.RuntimeException: No-args constructor for interface com.atg.lp.gson.Animal does not exist. Register an InstanceCreator with Gson for this type to fix this problem.

我明白为什么会发生这种情况，但我很难弄清楚处理这个问题的正确模式。我确实查看了指南，但它没有直接解决这个问题。

答案 1

下面是一个通用解决方案，适用于仅静态已知接口的所有情况。

Create serialiser/deserialiser：

final class InterfaceAdapter<T> implements JsonSerializer<T>, JsonDeserializer<T> {
    public JsonElement serialize(T object, Type interfaceType, JsonSerializationContext context) {
        final JsonObject wrapper = new JsonObject();
        wrapper.addProperty("type", object.getClass().getName());
        wrapper.add("data", context.serialize(object));
        return wrapper;
    }

    public T deserialize(JsonElement elem, Type interfaceType, JsonDeserializationContext context) throws JsonParseException {
        final JsonObject wrapper = (JsonObject) elem;
        final JsonElement typeName = get(wrapper, "type");
        final JsonElement data = get(wrapper, "data");
        final Type actualType = typeForName(typeName); 
        return context.deserialize(data, actualType);
    }

    private Type typeForName(final JsonElement typeElem) {
        try {
            return Class.forName(typeElem.getAsString());
        } catch (ClassNotFoundException e) {
            throw new JsonParseException(e);
        }
    }

    private JsonElement get(final JsonObject wrapper, String memberName) {
        final JsonElement elem = wrapper.get(memberName);
        if (elem == null) throw new JsonParseException("no '" + memberName + "' member found in what was expected to be an interface wrapper");
        return elem;
    }
}

让Gson将其用于您选择的接口类型：

Gson gson = new GsonBuilder().registerTypeAdapter(Animal.class, new InterfaceAdapter<Animal>())
                             .create();

答案 2

将动物设置为，则不会对其进行序列化。transient

或者你可以通过实现和（我认为这就是他们的名字......）自己序列化它。defaultWriteObject(...)defaultReadObject(...)

编辑请参阅此处有关“编写实例创建程序”的部分。

Gson 不能反序列化接口，因为它不知道将使用哪个实现类，因此您需要为 Animal 提供实例创建者并设置默认值或类似值。