如何为Gson编写自定义JSON反序列化器？

Question

如何为Gson编写自定义JSON反序列化器？

57

我有一个Java类，叫做User：

public class User
{
    int id;
    String name;
    Timestamp updateDate;
}

我从一个 Web 服务接收到一个包含用户对象的 JSON 列表：

[{"id":1,"name":"Jonas","update_date":"1300962900226"},
{"id":5,"name":"Test","date_date":"1304782298024"}]

我尝试编写了一个自定义的反序列化程序：

@Override
public User deserialize(JsonElement json, Type type,
                        JsonDeserializationContext context) throws JsonParseException {

        return new User(
            json.getAsJsonPrimitive().getAsInt(),
            json.getAsString(),
            json.getAsInt(),
            (Timestamp)context.deserialize(json.getAsJsonPrimitive(),
            Timestamp.class));
}

但是我的反序列化器不起作用。我该如何编写Gson的自定义JSON反序列化器？

- Jonas

这个例子有帮助吗？https://dev59.com/x2025IYBdhLWcg3wtoQ_ - Matt Ball

你说的“不工作”是什么意思？出了什么问题？ - ColinD

1

“date_date” 应该改为 “update_date” 吗？ - Programmer Bruce

3个回答

58

@Override
public User deserialize(JsonElement json, Type type,
        JsonDeserializationContext context) throws JsonParseException {

    JsonObject jobject = json.getAsJsonObject();

    return new User(
            jobject.get("id").getAsInt(), 
            jobject.get("name").getAsString(), 
            new Timestamp(jobject.get("update_date").getAsLong()));
}

我假设User类有适当的构造函数。

- Peter Knego

1

这种方法违背了使用GSON的初衷。 - SMR

我同意，它应该改为return context.deserialize(json, User.class);。 - nilesh

1

今天我在寻找一个东西，因为我的课程中有 java.time.Instant，而默认的gson无法反序列化它。我的POJOs看起来像这样：

open class RewardResult(
  @SerializedName("id")
  var id: Int,
  @SerializedName("title")
  var title: String?,
  @SerializedName("details")
  var details: String?,
  @SerializedName("image")
  var image: String?,
  @SerializedName("start_time")
  var startTimeUtcZulu: Instant?,   // Unit: Utc / Zulu. Unit is very important
  @SerializedName("end_time")
  var endTimeUtcZulu: Instant?,
  @SerializedName("unlock_expiry")
  var unlockExpiryTimeUtcZulu: Instant?,
  @SerializedName("target")
  var target: Int,
  @SerializedName("reward")
  var rewardItem: RewardItem
);

data class RewardItem(
  @SerializedName("type")
  var type: String?,
  @SerializedName("item_id")
  var itemId: Int,
  @SerializedName("amount")
  var amount: Int
)

然后对于Instant变量，我解析JSON的时间变量并将字符串转换为Instant。对于整数、字符串等，我使用jsonObject.get("id").asInt等方法。对于其他POJO，我使用默认的反序列化程序，如下所示：

val rewardItem: RewardItem = context!!.deserialize(rewardJsonElement,
        RewardItem::class.java);

因此，相应的自定义反序列化程序如下所示：

  val customDeserializer: JsonDeserializer<RewardResult> = object : JsonDeserializer<RewardResult> {
    override fun deserialize(json: JsonElement?, typeOfT: Type?, context: JsonDeserializationContext?): RewardResult {
      val jsonObject: JsonObject = json!!.asJsonObject;

      val startTimeString: String? = jsonObject.get("start_time")?.asString;
      var startTimeUtcZulu: Instant? = createTimeInstant(startTimeString);


      val endTimeString: String? = jsonObject.get("end_time")?.asString;
      var endTimeUtcZulu: Instant? = createTimeInstant(endTimeString);

      val unlockExpiryStr: String? = jsonObject.get("unlock_expiry")?.asString;
      var unlockExpiryUtcZulu: Instant? = createTimeInstant(unlockExpiryStr);

      val rewardJsonElement: JsonElement = jsonObject.get("reward");
      val rewardItem: ridmik.one.modal.reward.RewardItem = context!!.deserialize(rewardJsonElement,
          ridmik.one.modal.reward.RewardItem::class.java);  // I suppose this line means use the default jsonDeserializer

      var output: ridmik.one.modal.reward.RewardResult = ridmik.one.modal.reward.RewardResult(
          id = jsonObject.get("id").asInt,
          title = jsonObject.get("title")?.asString,
          details = jsonObject.get("details")?.asString,
          image = jsonObject.get("image")?.asString,
          startTimeUtcZulu = startTimeUtcZulu,
          endTimeUtcZulu = endTimeUtcZulu,
          unlockExpiryTimeUtcZulu = unlockExpiryUtcZulu,
          target = jsonObject.get("target").asInt,
          rewardItem = rewardItem
      );

      Timber.tag(TAG).e("output = "+output);

      return output;
    }

  }

最终，我像这样创建了自定义的Gson：

 val gsonBuilder = GsonBuilder();
 gsonBuilder.registerTypeAdapter(RewardResult::class.javaObjectType,
          this.customJsonDeserializer);
      val customGson: Gson = gsonBuilder.create();

- Qazi Fahim Farhan

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- Programmer Bruce · Accepted Answer

我会采用稍微不同的方法，以尽量减少代码中的“手动”解析，否则就会有点违背我使用像Gson这样的API的初衷。

// output:
// [User: id=1, name=Jonas, updateDate=2011-03-24 03:35:00.226]
// [User: id=5, name=Test, updateDate=2011-05-07 08:31:38.024]

// using java.sql.Timestamp

public class Foo
{
  static String jsonInput = 
    "[" +
      "{\"id\":1,\"name\":\"Jonas\",\"update_date\":\"1300962900226\"}," +
      "{\"id\":5,\"name\":\"Test\",\"update_date\":\"1304782298024\"}" +
    "]";

  public static void main(String[] args)
  {
    GsonBuilder gsonBuilder = new GsonBuilder();
    gsonBuilder.setFieldNamingPolicy(FieldNamingPolicy.LOWER_CASE_WITH_UNDERSCORES);
    gsonBuilder.registerTypeAdapter(Timestamp.class, new TimestampDeserializer());
    Gson gson = gsonBuilder.create();
    User[] users = gson.fromJson(jsonInput, User[].class);
    for (User user : users)
    {
      System.out.println(user);
    }
  }
}

class User
{
  int id;
  String name;
  Timestamp updateDate;

  @Override
  public String toString()
  {
    return String.format(
      "[User: id=%1$d, name=%2$s, updateDate=%3$s]",
      id, name, updateDate);
  }
}

class TimestampDeserializer implements JsonDeserializer<Timestamp>
{
  @Override
  public Timestamp deserialize(JsonElement json, Type typeOfT, JsonDeserializationContext context)
      throws JsonParseException
  {
    long time = Long.parseLong(json.getAsString());
    return new Timestamp(time);
  }
}

(这里假设原问题中的"date_date"应该是"update_date")