Question

我可以将这个复杂的json转换为java类吗？我正在使用GSON来解析JSON数据。

http://api.feedzilla.com/v1/categories/26/articles.json

这样的事情：

Parsing a complex Json Object using GSON in Java

编辑：我在JAVA中使用了GSON的这个例子复杂JSON对象

这是我的课文：

public class Article {
public String publish_date;
public String source;
public String source_url;
public String summary;
public String title;
public String url;

}

DataStorage类：

public class DataStorage {
public static List<Article> article;

}

MyActivity类：

public class MyActivity extends Activity {
/**
 * Called when the activity is first created.
 */
@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.main);

    AsyncHttpClient httpClient = new AsyncHttpClient();

    httpClient.get("http://api.feedzilla.com/v1/categories/26/articles.json", null,
            new JsonHttpResponseHandler()
            {
                @Override
                public void onSuccess(int statusCode, Header[] headers, JSONArray response)
                {
                    Log.d("HTTP_RESPONSE", response.toString());

                    Gson myGson = new Gson();
                    DataStorage ac = myGson.fromJson(response.toString(), DataStorage.class);

                    Log.d("HTTP_RESPONSE", response.toString());

                    //listView.setAdapter(new Adapter(getApplicationContext()));

                }
            }
    );
}

}

我编译后在logcat中显示，为什么？

12-30 16:22:20.044    1156-1156/com.example.Articles W/JsonHttpResponseHandler﹕ onSuccess(int, Header[], JSONObject) was not overriden, but callback was received

Answer 1

使用JSON美化器（如http://jsonformatter.curiousconcept.com/），这样您就可以更好地了解其中的内容。然后，创建类，如下所示：

class article {
    String publish_date;
    String title;
    ...

复杂的JSON到Java类

1 个答案: