問題接受

Question

昨天它工作正常，今天我得到這樣的消息：java.lang.String類型的值不能轉換爲JSONArray ....

這是

private class HttpAsyncTaskFamilies extends AsyncTask<String, Void, String> { 

@Override 
protected String doInBackground(String... urls) { 
    return GET(urls[0]); 
} 

// onPostExecute displays the results of the AsyncTask 
@Override 
protected void onPostExecute(String result) { 
    JSONArray jArray; 
    String families = ""; 
    Log.d(TAG, "result in onPostExectute Families -> " + result); 
    try { 
     DatabaseHelper helper = new DatabaseHelper(
       getApplicationContext()); 

     try { 
      helper.dropAndCreate(helper.getConnectionSource(), 
        Familia.class); 

     // Do the inserts in table Familia 
       jArray = new JSONArray(result); 

       Log.d(TAG, "jsonArray -> " + jArray); 
       for (int i = 0; i < jArray.length(); i++) { 
        JSONObject json = jArray.getJSONObject(i); 
        Log.d(TAG, "jsonObject -> " + json.toString()); 

        String id = json.getString("familia_id"); 
        Log.d(TAG, "id -> " + id); 
        int id_family = Integer.parseInt(json 
          .getString("familia_id")); 

        // for each Familia 
        Familia familia = new Familia(json.getInt("familia_id"), 
          json.getString("nom_familia")); 
        helper.getFamiliaDao().create(familia); 
        families += json.getString("nom_familia"); 
        Log.d(TAG, "Familia inserida -> " + familia); 
       } 
       Log.d(TAG, "families actualitzades! " + families); 

      } catch (SQLException e) { 
       e.printStackTrace(); 
      } 
     } catch (JSONException e) { 
      e.printStackTrace(); 
     } 

    } 
} 

的錯誤是在行：那我收到的JSON：當我嘗試從做一個JSONArray

和代碼將停止： jArray =新的JSONArray（結果）;

任何想法？這是奇怪的是，昨天工作正常，今天我只能有這個錯誤...

找不到方法Add：

更多信息：

在onCreate方法我有：

new HttpAsyncTaskFamilies（）.execute（「http://aplicacionesmovilandroid.es/json/families.php」）;

然後當HttpAsyncTaskFamilies（）調用GET方法，這是一個方法：

public String GET(String url) { 
InputStream inputStream = null; 
String result = ""; 
try { 
    // create HttpClient 
    HttpClient httpClient = new DefaultHttpClient(); 

    UtilDAOImpl dao = new UtilDAOImpl(this); 

     Versio versioObject = dao.lookupVersioCentre("centres"); 
     String versio = versioObject.getVersio(); 

     // we change the white space for %20, a representation of the white 
     // space 
     // that php can recognise. Without it, the php file always deceive a 
     // null value 
     // instead of the actual version 
     // String finalVersion = versio.replace(" ", "%20"); 

     // make GET request to the given URL 
     // HttpResponse httpResponse = httpClient.execute(new HttpGet(url 
     // + "?centres_versio=" + finalVersion)); 

     HttpResponse httpResponse = httpClient.execute(new HttpGet(url)); 

     // receive response as InputStream 
     inputStream = httpResponse.getEntity().getContent(); 

     // convert InputStream to string 
     if (inputStream != null) { 
      result = convertInputStreamToString(inputStream); 
     } else { 
      result = "No ha funcionat!"; 
     } 
    } catch (Exception e) { 
     Log.d("InputStream", e.getLocalizedMessage()); 
    } 
    return result; 
} 

Answer 1

得到響應後，你必須result轉化爲JSONObject如..

JSONObject jsonObj = new JSONObject(result); 
FamiliesArray = jsonObj.getJSONArray(Families); 

可以和我看到你錯過了..

Answer 2

你的代碼沒有錯，但是 這些行可能會產生問題 還是沒法確定！

評論這些行！

Log.d(TAG,"arrayJson ->"+arrayJson.toString()); 

Answer 3

替換你的代碼，

Familia familia = new Familia(json.getInt("familia_id"), 
         json.getString("nom_familia")); 

到

Familia familia = new Familia(Integer.parseInt(json.getString("familia_id")), 
         json.getString("nom_familia")); 

因爲服務器發送familia_id是String形式。希望這項工作...... :)

Answer 4

你所指的代碼似乎是正確的。你可以修剪字符串，並嘗試將修剪後的字符串傳遞給JSONArray的構造函數嗎？

另外，我會建議使用傑克遜分析器將JSON轉換爲類對象）。例子可以在這裏找到，http://programmerbruce.blogspot.in/2011/05/deserialize-json-with-jackson-into.html