如何檢查url是否存在或取消json數據

我正在提取不同book_name的json數據&其工作正常。但如果書不存在，我的應用程序崩潰。我知道JsonObject獲取空數據，但我當書未找到時，希望它做某件事而不是崩潰。我的代碼是： `如何檢查url是否存在或取消json數據

public class AsyncTaskParseJson extends AsyncTask<String, String, String> { 

    final String TAG = "AsyncTaskParseJson.java"; 

    // set your json string url here 
    Intent po=getIntent(); 
    String bok=po.getStringExtra("bookname"); 
    String newName = bok.replaceAll("\\s", "%20"); 
    String yourJsonStringUrl = "http://www.bsservicess.com/photoUpload/star_avg.php?bookName="+newName; 

    // contacts JSONArray 
    JSONArray dataJsonArr = null; 
    @Override 
    protected void onPreExecute() {} 


    protected String doInBackground(String... arg0) { 
     if(exists(yourJsonStringUrl)) { 
     try { 
      JSONParser jParser = new JSONParser(); 

      // get json string from url 

      JSONObject json = jParser.getJSONFromUrl(yourJsonStringUrl); 



       // get the array of users 
       dataJsonArr = json.getJSONArray("result"); 
       JSONObject c = dataJsonArr.getJSONObject(0); 
       // na=c.getString("avg"); 

       ratenumS = c.getString("avg"); 
       numberS = c.getString("num"); 
       starts = Float.parseFloat(c.getString("avg")); 


     } catch (JSONException e) { 
      e.printStackTrace(); 
     } 
     }else{ 
      starts=0; 
      numberS="0"; 
     } 

     return null; 
    } 

    @Override 
    protected void onPostExecute(String strFromDoInBg) { 
     super.onPostExecute(strFromDoInBg); 
     ratenum.setText(ratenumS); 
     number.setText("("+numberS+")"); 
     netRate.setRating(starts); 
    // Toast.makeText(mybookview.this, Float.toString(starts),Toast.LENGTH_SHORT).show(); 
    }

Plz幫助我。 `

 public JSONObject getJSONFromUrl(String url) { 

    // make HTTP request 
    try { 

     DefaultHttpClient httpClient = new DefaultHttpClient(); 
     HttpPost httpPost = new HttpPost(url); 

     HttpResponse httpResponse = httpClient.execute(httpPost); 
     HttpEntity httpEntity = httpResponse.getEntity(); 
     is = httpEntity.getContent(); 

    } catch (UnsupportedEncodingException e) { 
     e.printStackTrace(); 
    } catch (ClientProtocolException e) { 
     e.printStackTrace(); 
    } catch (IOException e) { 
     e.printStackTrace(); 
    } 

    try { 

     BufferedReader reader = new BufferedReader(new InputStreamReader(is, "iso-8859-1"), 8); 
     StringBuilder sb = new StringBuilder(); 
     String line = null; 
     while ((line = reader.readLine()) != null) { 
      sb.append(line + "\n"); 
     } 
     is.close(); 
     json = sb.toString(); 

    } catch (Exception e) { 
     Log.e(TAG, "Error converting result " + e.toString()); 
    } 

    // try parse the string to a JSON object 
    try { 
     jObj = new JSONObject(json); 
    } catch (JSONException e) { 
     Log.e(TAG, "Error parsing data " + e.toString()); 
    } 

    // return JSON String 
    return jObj; 
}

來源

2016-07-15 Nicky Manali

你可以發佈你的'JSONParser'類嗎？ – Bill

okk我發表在mty編輯問題 –

// get json string from url 

JSONObject json = jParser.getJSONFromUrl(yourJsonStringUrl); 
if(json == null){ 
    //url parsing didn't work 
    return; 
}

來源

2016-07-15 18:57:46 Bill

大於thanxx來解決我的問題。 –

您需要處理外，加試捕在您的應用程序崩潰。

來源

2016-07-15 18:44:57 KgaboL

我知道它在哪裏崩潰引起：java.lang.NullPointerException：嘗試調用虛擬方法'org.json.JSONArray org.json.JSONObject.getJSONArray（java.lang.String）'在空對象引用 at dataJsonArr = json.getJSONArray（「result」）;但dnt知道如何糾正它 –

json null後... JSONObject json = jParser.getJSONFromUrl（yourJsonStringUrl）; ？如果是，那麼你需要一個空檢查，然後再嘗試使數組 – KgaboL

如何檢查url是否存在或取消json數據

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