Android：从AsyncTask返回

Question

请注意：我已阅读其他相关问题。

我有一个DatabaseCall类，其中包含检索特定数据的函数。我已经在下面添加了代码，我只是得到了所有问题等。

我将从testActivity.class拨打此邮件，因此我可以填充list<Questions>，但是因为这是异步的，我无法从onPostExecute返回任何内容; d在活动中称之为 - 但是我的代码变得太乱了。

以下CodeCall类代码

public class DatabaseCalls {

public List<Question> getTestQuestionsFromDatabase(){
    class LoginAsync extends AsyncTask<String, Void, String>{

        private Dialog loading;

        @Override
        protected List<Question> onPostExecute(String s){
            loading.dismiss();
            Log.d("PostExecute", "JSON!: " + s);
            try{
                JSONObject jObj = new JSONObject(s);
                boolean error = jObj.getBoolean("error");

                List<Question> listQuestions = new ArrayList<Question>();

                if (!error){ 
                    for (int i=0; i<11; i++){
                        JSONObject question = jObj.getJSONObject(Integer.toString(i));
                        Question qObj = new Question(Integer.parseInt(question.getString("questionID")), question.getString("question"), Integer.parseInt(question.getString("answerID")), Integer.parseInt(question.getString("complexityID")));
                        listQuestions.add(qObj);                                    
                    }     
                    return listQuestions;
                }else{
                     String errorMsg = jObj.getString("error_msg");
                     /**
                      * Can't toast here because it is a helper class - it isn't
                      * an activity - need to return an empty object instead?
                      */
                     //Toast.makeText(getApplicationContext(),errorMsg, Toast.LENGTH_LONG).show();
                }
            }catch(JSONException e){
                e.printStackTrace();
            }               
        }

        @Override
        protected String doInBackground(String... params) {
            String s = params[0];
            BufferedReader bufferedReader = null;
            try {
                URL url = new URL(AppConfig.Questions_URL + s);
                HttpURLConnection con = (HttpURLConnection) url.openConnection();
                con.setRequestProperty ("Authorization", "Basic Z2FycmV0dGg6ZnJBc3Rpbmc0");
                bufferedReader = new BufferedReader(new InputStreamReader(con.getInputStream()));
                //This string is returned to the onPostExecute
                String result;                    
                result = bufferedReader.readLine();
                return result;
            }catch(Exception e){
                Log.d("Exception in try", "Exception" + e.toString());
                return null;
            }
        }           
    } 
    LoginAsync la = new LoginAsync();
    la.execute();
}}

我该如何对抗这个？并返回一个值。