用GSON序列化带有通用嵌套数据对象的响应对象

Question

我正在使用带有GSON的改进型来实现JSON序列化，并使用Realm来实现存储。

{
   status : 1
   data: { object data }
}

public class ResponseDeserializer implements JsonDeserializer {

@Override
public Object1 deserialize(JsonElement je, Type type, JsonDeserializationContext jdc)
{
    JsonElement data = je.getAsJsonObject().get("data");
    GsonBuilder gsonBuilder = new GsonBuilder()
            .setExclusionStrategies(new ExclusionStrategy() {
                @Override
                public boolean shouldSkipField(FieldAttributes f) {
                    return f.getDeclaringClass().equals(RealmObject.class);
                }

                @Override
                public boolean shouldSkipClass(Class<?> clazz) {
                    return false;
                }
            });
    Gson gson = gsonBuilder.create();
    Object1 o1 = gson.fromJson(data, new TypeToken<Object1>() {}.getType());
    return o1;
}
}

第一个问题：当我尝试将fromJson更改为泛型类型时，会产生一个带有对象键值对的“com.google.gson.internal.LinkedTreeMap”，代码：

public class ResponseDeserializer<T> implements JsonDeserializer<T> {

@Override
public T deserialize(JsonElement je, Type type, JsonDeserializationContext jdc)
{
    JsonElement data = je.getAsJsonObject().get("data");
    GsonBuilder gsonBuilder = new GsonBuilder()
            .setExclusionStrategies(new ExclusionStrategy() {
                @Override
                public boolean shouldSkipField(FieldAttributes f) {
                    return f.getDeclaringClass().equals(RealmObject.class);
                }

                @Override
                public boolean shouldSkipClass(Class<?> clazz) {
                    return false;
                }
            });
    Gson gson = gsonBuilder.create();
    T o = gson.fromJson(data, new TypeToken<T>() {}.getType());
    return o;
}
}

第二个问题：反序列化器如何根据json结果返回对象或对象列表。

Answer 1

现在我用一种不同的方法解决了它：

2列表和对象响应模型：

public class ApiResponse<T> {
    private int success;
    private String[] errors;
    private String[] messages;
    private T data;
}

public class ApiListResponse<T> {
    private int success;
    private String[] errors;
    private String[] messages;
    private List<T> data;
}

1响应模型的反序列化器

public class ResponseDeserializer<T> implements JsonDeserializer {
@Override
public T deserialize(JsonElement je, Type type, JsonDeserializationContext jdc) throws JsonParseException {
    JsonElement data = je.getAsJsonObject().get("data");
    GsonBuilder gsonBuilder = new GsonBuilder()
            .setExclusionStrategies(new ExclusionStrategy() {
                @Override
                public boolean shouldSkipField(FieldAttributes f) {
                    return f.getDeclaringClass().equals(RealmObject.class);
                }

                @Override
                public boolean shouldSkipClass(Class<?> clazz) {
                    return false;
                }
            });
    Gson gson = gsonBuilder.create();
    T response = gson.fromJson(je, type);
    return response;
}
}

现在我可以为GSON注册类型，比如：

gsonBuilder.registerTypeAdapter(new TypeToken<ApiResponse<Object1>>() {}.getType(), new ResponseDeserializer<ApiResponse>());
gsonBuilder.registerTypeAdapter(new TypeToken<ApiListResponse<Object1>>() {}.getType(), new ResponseDeserializer<ApiListResponse>());

现在将正确转换以下JSON响应：

API响应

{  
   "data":{  
      "Object1Id":"1",
      "Name":"Test 1",
   },
   "messages":[  
   ],
   "errors":[  
   ],
   "success":"1"
}

ApilistResponse

{  
   "data":[{  
      "Object1Id":"1",
      "Name":"Test 1",
   },{  
      "Object1Id":"2",
      "Name":"Test 2",
   }],
   "messages":[  
   ],
   "errors":[  
   ],
   "success":"1"
}

有更好的解决方案吗？