獲得圖像名稱CodeNameOne

Question

我想從數據庫 對象的列表我是100％，我retreive數據，但名單，所以我的PHP代碼似乎是很好，當我想

public ArrayList<Categorie> getListCategorie() { 
    ArrayList<Categorie> listcategories = new ArrayList<>(); 


    ConnectionRequest con2 = new ConnectionRequest(); 
    con2.setUrl("http://localhost/pidev2017/selectcategorie.php"); 
    con2.addResponseListener(new ActionListener<NetworkEvent>() { 
     @Override 
     public void actionPerformed(NetworkEvent evt) { 
     try { 
     JSONParser j = new JSONParser(); 
     Map<String, Object> catefories = j.parseJSON(new CharArrayReader(new String(con2.getResponseData()).toCharArray())); 
     List<Map<String, Object>> list = (List<Map<String, Object>>) catefories.get("Categorie"); 
     for (Map<String, Object> obj : list) { 
      Categorie categorie = new Categorie(); 
      categorie.setId(Integer.parseInt(obj.get("id").toString())); 
      categorie.setNomCategorie(obj.get("nomCategorie").toString()); 
      listcategories.add(categorie); 
     } 
    } catch (IOException ex) { 
    } 
     } 

    }); 
    NetworkManager.getInstance().addToQueue(con2); 
    return listcategories; 
} 

獲取我的結果 「listcategories」 我發現是空

Answer 1

變化

NetworkManager.getInstance().addToQueue(con2); 

到

NetworkManager.getInstance().addToQueueAndWait(con2); 

您可能會在獲取數據之前嘗試獲取結果。