Json 使用 GSON 反序列化泛型类型

在反序列化时，必须指定T的类型.如果Gson不知道要实例化哪一个Type，那么如何创建posts中的List？它不可能永远保持T.因此，您可以提供类型T作为Class参数.

Now assuming, the type of posts was String you would deserialize MyJson<String> as (I've also added a String json parameter for simplicity; you would read from your reader as before):

doInBackground(String.class, "{posts: [\"article 1\", \"article 2\"]}");

protected MyJson<T> doInBackground(Class<T> type, String json, Void... params) {

    GsonBuilder gson = new GsonBuilder();
    Type collectionType = new TypeToken<MyJson<T>>(){}.getType();

    MyJson<T> myJson = gson.create().fromJson(json, collectionType);

    System.out.println(myJson.getPosts()); // ["article 1", "article 2"]
    return myJson;
}

类似地，反序列化Boolean个对象中的MyJson个

doInBackground(Boolean.class, "{posts: [true, false]}");

protected MyJson<T> doInBackground(Class<T> type, String json, Void... params) {

    GsonBuilder gson = new GsonBuilder();
    Type collectionType = new TypeToken<MyJson<T>>(){}.getType();

    MyJson<T> myJson = gson.create().fromJson(json, collectionType);

    System.out.println(myJson.getPosts()); // [true, false]
    return myJson;
}

I've assumed MyJson<T> for my examples to be as

public class MyJson<T> {

    public List<T> posts;

    public List<T> getPosts() {
        return posts;
    }
}

所以，如果你想反序列化一个List<MyObject>，你可以调用这个方法

// assuming no Void parameters were required
MyJson<MyObject> myJson = doInBackground(MyObject.class);