Question

我有以下课程

public class MyClass {
    private List<MyOtherClass> others;
}

public class MyOtherClass {
    private String name;
}

我有JSON可能看起来像这样

{
  others: {
    name: "val"
  }
}

或者

{
  others: [
    {
      name: "val"
    },
    {
      name: "val"
    }
  ]
}

我希望能够对这两种JSON格式使用相同的MyClass。有没有办法用Gson做到这一点？

Answer 1

我想出了一个答案。

private static class MyOtherClassTypeAdapter implements JsonDeserializer<List<MyOtherClass>> {
    public List<MyOtherClass> deserialize(JsonElement json, Type typeOfT, JsonDeserializationContext ctx) {
        List<MyOtherClass> vals = new ArrayList<MyOtherClass>();
        if (json.isJsonArray()) {
            for (JsonElement e : json.getAsJsonArray()) {
                vals.add((MyOtherClass) ctx.deserialize(e, MyOtherClass.class));
            }
        } else if (json.isJsonObject()) {
            vals.add((MyOtherClass) ctx.deserialize(json, MyOtherClass.class));
        } else {
            throw new RuntimeException("Unexpected JSON type: " + json.getClass());
        }
        return vals;
    }
}

实例化像这样的Gson对象

Type myOtherClassListType = new TypeToken<List<MyOtherClass>>() {}.getType();

Gson gson = new GsonBuilder()
        .registerTypeAdapter(myOtherClassListType, new MyOtherClassTypeAdapter())
        .create();

TypeToken是com.google.gson.reflect.TypeToken。

您可以在此处阅读解决方案：

https://sites.google.com/site/gson/gson-user-guide#TOC-Serializing-and-Deserializing-Gener

Answer 2

谢谢三杯的解决方案！

如果需要多种类型，则使用泛型类型：

public class SingleElementToListDeserializer<T> implements JsonDeserializer<List<T>> {

private final Class<T> clazz;

public SingleElementToListDeserializer(Class<T> clazz) {
    this.clazz = clazz;
}

public List<T> deserialize(JsonElement json, Type typeOfT, JsonDeserializationContext context) throws JsonParseException {
    List<T> resultList = new ArrayList<>();
    if (json.isJsonArray()) {
        for (JsonElement e : json.getAsJsonArray()) {
            resultList.add(context.<T>deserialize(e, clazz));
        }
    } else if (json.isJsonObject()) {
        resultList.add(context.<T>deserialize(json, clazz));
    } else {
        throw new RuntimeException("Unexpected JSON type: " + json.getClass());
    }
    return resultList;
    }
}

配置Gson：

Type myOtherClassListType = new TypeToken<List<MyOtherClass>>() {}.getType();
SingleElementToListDeserializer<MyOtherClass> adapter = new SingleElementToListDeserializer<>(MyOtherClass.class);
Gson gson = new GsonBuilder()
    .registerTypeAdapter(myOtherClassListType, adapter)
    .create();

Answer 3

建立三杯答案，我有以下内容让JsonArray直接反序列化为数组。

static public <T> T[] fromJsonAsArray(Gson gson, JsonElement json, Class<T> tClass, Class<T[]> tArrClass)
        throws JsonParseException {

    T[] arr;

    if(json.isJsonObject()){
        //noinspection unchecked
        arr = (T[]) Array.newInstance(tClass, 1);
        arr[0] = gson.fromJson(json, tClass);
    }else if(json.isJsonArray()){
        arr = gson.fromJson(json, tArrClass);
    }else{
        throw new RuntimeException("Unexpected JSON type: " + json.getClass());
    }

    return arr;
}

用法：

    String response = ".......";

    JsonParser p = new JsonParser();
    JsonElement json = p.parse(response);
    Gson gson = new Gson();
    MyQuote[] quotes = GsonUtils.fromJsonAsArray(gson, json, MyQuote.class, MyQuote[].class);

Answer 4

共享代码，并将反序列化逻辑仅应用于特定字段：

JSON模型：

public class AdminLoginResponse implements LoginResponse
{
    public Login login;
    public Customer customer;
    @JsonAdapter(MultiOrganizationArrayOrObject.class)    // <-------- look here
    public RealmList<MultiOrganization> allAccounts;
}

抽象类：

/**
 * parsed field can either be a [JSONArray] of type [Element], or an single [Element] [JSONObject].
 */
abstract class ArrayOrSingleObjectTypeAdapter<TypedList: List<Element>, Element : Any>(
    private val elementKClass: KClass<Element>
) : JsonDeserializer<TypedList> {
    override fun deserialize(
        json: JsonElement, typeOfT: Type?, ctx: JsonDeserializationContext
    ): TypedList = when {
        json.isJsonArray -> json.asJsonArray.map { ctx.deserialize<Element>(it, elementKClass.java) }
        json.isJsonObject -> listOf(ctx.deserialize<Element>(json, elementKClass.java))
        else -> throw RuntimeException("Unexpected JSON type: " + json.javaClass)
    }.toTypedList()
    abstract fun List<Element>.toTypedList(): TypedList
}

特定于字段的类：

class MultiOrganizationArrayOrObject
    : ArrayOrSingleObjectTypeAdapter<RealmList<MultiOrganization>,MultiOrganization>(kClass()) {
    override fun List<MultiOrganization>.toTypedList() = RealmList(*this.toTypedArray())
}

Gson处理对象或数组

4 个答案: