如何反序列化作为单个对象或带有GSON的对象列表传递的字段？ [英] Howto deserialize field that is passed either as a single Object or as a list of Objects with GSON?

问题描述

我已成功使用GSON将JSON转换为单个Object并将JSON转换为对象列表。我的问题是有2个源向我发送数据。一个是发送一个对象，另一个是发送一个对象列表。

I have successfully used GSON to convert a JSON to a single Object and to convert JSON to a list of objects. My problem is that there are 2 sources emitting data to me. One is only sending one object and the other is sending a list of Objects.

单个对象来自第一个来源：

{
    id : '1',
    title: 'sample title',
    ....
}

来自第二来源的对象列表：

[
    {
        id : '1',
        title: 'sample title',
        ....
    },
    {
        id : '2',
        title: 'sample title',
        ....
    },
    ...
]

班级用于反序列化：

 public class Post {

      private String id;
      private String title;

      /* getters & setters */
 }

以下是我的第一个案例：

Below is working for my 1st case:

Post postData = gson.fromJson(jsonObj.toString(), Post.class);

这适用于我的第二种情况：

And this is working for my 2nd case:

Post[] postDatas = gson.fromJson(jsonObj.toString(), Post[].class);

有没有办法管理这两种情况？或者我应该查看字符串并在不可用时添加[]
谢谢

Is there a way to manage both cases? Or should I look into the string and add [] when it is not available Thanks

推荐答案

如何创建一个自定义反序列化器，用于检查json是否为数组，如果没有，则创建一个包含单个对象的数组，如：

How about creating a custom deserializer that checks if the json is an array and if not creates an array having the single object in it, like:

public class PostArrayOrSingleDeserializer implements JsonDeserializer<Post[]> {

    private static final Gson gson = new Gson();

    public Post[] deserialize(JsonElement json, Type typeOfT, 
                JsonDeserializationContext ctx) {
        try {
            return gson.fromJson(json.getAsJsonArray(), typeOfT);
        } catch (Exception e) {
            return new Post[] { gson.fromJson(json, Post.class) };
        }
    }
}

并将其添加到您的Gson ：

and adding it to your Gson:

Post[] postDatas = new GsonBuilder().setPrettyPrinting()
    .registerTypeAdapter(Post[].class, new PostArrayOrSingleDeserializer())
    .create()
    .fromJson(jsonObj.toString(), Post[].class);

因此，您应始终拥有 Post 包含一件或多件物品。

So as a result you should always have an array of Post with on or more items.

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