2017-07-18 11 views
0

下面我試圖創建一個列表,但符號「Feed」無法解析。我應該包括哪些內容。這個想法是從我創建的文件中獲取新聞文章。無法使用Json爲我的列表解析符號

public class Utils { 
    private static final String TAG = "Utils"; 
    public static List<Feed> loadFeeds(Context context){ 
     try{ 
      GsonBuilder builder = new GsonBuilder(); 
      Gson gson = builder.create(); 
      JSONArray array = new JSONArray(loadJSONFromAsset(context, "news.json")); 
      List<Feed> feedList = new ArrayList<>(); 
      for(int i=0;i<array.length();i++){ 
       Feed feed = gson.fromJson(array.getString(i), Feed.class); 
       feedList.add(feed); 
      } 
      return feedList; 
     }catch (Exception e){ 
      Log.d(TAG,"seedGames parseException " + e); 
      e.printStackTrace(); 
      return null; 
     } 
    } 
    private static String loadJSONFromAsset(Context context, String jsonFileName) { 
     String json = null; 
     InputStream is = null; 
     try { 
      AssetManager manager = context.getAssets(); 
      Log.d(TAG,"path "+jsonFileName); 
      is = manager.open(jsonFileName); 
      int size = is.available(); 
      byte[] buffer = new byte[size]; 
      is.read(buffer); 
      is.close(); 
      json = new String(buffer, "UTF-8"); 
     } catch (IOException ex) { 
      ex.printStackTrace(); 
      return null; 
     } 
     return json; 
    } 
} 

*

回答

0

要創建List<Feed>,你必須創建Feed類並將其導入到Utils類。

+0

Got sorted。如果我的Feed.java類拼寫錯誤。請原諒我笨屁股。謝謝 –

0

已排序。如果我的Feed.java類拼寫錯誤。請原諒我的啞巴屁股