如何使用Gson动态处理json响应数组/对象

时间:2013-01-17 06:37:03

标签: java android gson

我遇到了一个问题,有时Json响应会返回一个对象数组,有时会反对自己,我们如何在响应类中动态处理。 在当前例如:结果有时会得到一个对象数组

 "\"results\": " +
            "[{" +

有时反对自己

 "\"results\": " +
            "{" +

例如:

我们如何处理这个问题?

Gson gson = new Gson();
    SearchResponse response=new SearchResponse();
    response= gson.fromJson("{" +
            "\"completed_in\": 0.047," +
            "\"max_id\": 291771567376039936," +
            "\"max_id_str\": \"291771567376039936\"," +
            "\"next_page\": \"?page=2&max_id=291771567376039936&q=javacodegeeks\"," +
            "\"page\": 1," +
            "\"query\": \"javacodegeeks\"," +
            "\"refresh_url\": \"?since_id=291771567376039936&q=javacodegeeks\"," +
            "\"results\": " +
            "{" +
            "\"created_at\": \"Thu, 17 Jan 2013 04:58:57 +0000\"," +
            "\"from_user\": \"hkokko\"," +
            "\"from_user_id\": 24726686," +
            "\"from_user_id_str\": \"24726686\"," +
            " \"from_user_name\": \"Hannu Kokko\"," +
            " \"geo\": null," +
            "\"id\": 291771567376039936," +
            "\"id_str\": \"291771567376039936\"," +
            "\"iso_language_code\": \"en\"," +
            " \"metadata\": {" +
            "\"result_type\": \"recent\"}," +
            "\"profile_image_url\": \"hjh\"," +
            "\"profile_image_url_https\": \"kkj\"," +
            "\"source\": \"<a href="hj;\"," +
            "\"text\": \"Continuous Deployment: Are You Afraid It Might Work? jh\"," +
            "\"to_user\": null," +
            "\"to_user_id\": 0," +
            "\"to_user_id_str\": \"0\"," +
            "\"to_user_name\": null" +
            " }," +
            "\"results_per_page\": 15," +
            "\"since_id\": 0," +
            "\"since_id_str\": \"0\"" +
            "}", SearchResponse.class);
    System.out.println(response.toString());

请帮助......

任何人都可以通过使用不同的罐子来提出任何建议吗?

2 个答案:

答案 0 :(得分:17)

我为此找到了一个解决方案,我觉得要分享这个......代码会自动转换..如果响应是响应类中的arraylist,那么如果对象响应,那么如果arraylist则加入arraylist else它将采用相同的列表。  我们需要挂钩更改它从json调用的响应。

public class ArrayAdapter<T> extends TypeAdapter<List<T>> {
    private Class<T> adapterclass;

    public ArrayAdapter(Class<T> adapterclass) {

        this.adapterclass = adapterclass;
    }

    public List<T> read(JsonReader reader) throws IOException {


        List<T> list = new ArrayList<T>();

        Gson gson = new Gson();

        if (reader.peek() == JsonToken.BEGIN_OBJECT) {

            T inning = (T) gson.fromJson(reader, adapterclass);
            list.add(inning);

        } else if (reader.peek() == JsonToken.BEGIN_ARRAY) {

            reader.beginArray();
            while (reader.hasNext()) {
                T inning = (T) gson.fromJson(reader, adapterclass);
                list.add(inning);
            }
            reader.endArray();

        } else {
            reader.skipValue();
        }

        return list;
    }

    public void write(JsonWriter writer, List<T> value) throws IOException {

    }

}

public class ArrayAdapterFactory implements TypeAdapterFactory {

  @SuppressWarnings({ "unchecked" })
  @Override
  public <T> TypeAdapter<T> create(final Gson gson, final TypeToken<T> type) {

      ArrayAdapter typeAdapter = null;
      try {
          if (type.getRawType() == List.class)
          {

              typeAdapter = new ArrayAdapter(
                      (Class) ((ParameterizedType) type.getType())
                              .getActualTypeArguments()[0]);
          }
      } catch (Exception e) {
          e.printStackTrace();
      }

      return typeAdapter;
}

然后打电话

 Gson gson = new GsonBuilder().registerTypeAdapterFactory(new ArrayAdapterFactory()).create();
 SearchResponse response;
 esponse= gson.fromJson("your json string", SearchResponse.class)

答案 1 :(得分:3)

您需要编写一个自定义反序列化程序,用于检查JSON中results的类型,然后相应地执行操作。

您的POJO将包含results的数组,如果您的传入JSON只有一个对象,则需要修复它。一种方法是修改JSON然后反序列化它:

class SearchResponseDeserializer implements JsonDeserializer<SearchResponse> {
  public SearchResponse deserialize(JsonElement json, Type typeOfT, JsonDeserializationContext context)
      throws JsonParseException {

    if (json.getAsJsonObject().get("results").isJsonObject()) {
      //modify JSON: change results to be an array
      // ...
    }

    return new Gson().fromJson(json, SearchResults.class);

  }
}

或者,当然要修复服务器。应始终返回一个数组以避免此问题。