不允许我们在SharedPreferences中存储cookie并从AsyncTask类返回JSONObject

Question

我试图将从服务器接收的cookie存储到共享首选项中，然后在扩展AsyncTask的类的重写doinbackground方法中返回JSONObject。

但这显示错误在哪里我检查onpostexecute方法的字符串即

if (jsonObject.getString("firstTime").equalsIgnoreCase("true")) 

，但是当我删除写入存储的cookie在sharedPreferences字符串中的代码，然后它工作正常，即声明

List<Cookie> cookies = httpClient.getCookieStore().getCookies(); 

      cook = cookies.toString(); 

      // to store only the value of cookie i.e 26 character long 
      cook = cook.substring(38, 64); 
      Editor editor = sharedPreferences.edit(); 
      editor.putString("cookie", result); 
      editor.commit(); 

这里是两种方法的完整代码。

@Override 
     protected JSONObject doInBackground(String... params) { 
      // TODO Auto-generated method stub 

      String cook = ""; 

      JSONObject jsonObject = null; 
      try { 

       List<BasicNameValuePair> list = new ArrayList<BasicNameValuePair>(); 
       list.add(new BasicNameValuePair("type", "login")); 
       list.add(new BasicNameValuePair("email", editEmail.getText() 
         .toString())); 
       list.add(new BasicNameValuePair("password", editPassword 
         .getText().toString())); 
       // list.add(new BasicNameValuePair("cookie", receivedCookie)); 
       DefaultHttpClient httpClient = new DefaultHttpClient(); 
       HttpPost httpPost = new HttpPost(params[0]); 
       httpPost.setEntity(new UrlEncodedFormEntity(list)); 

       // executing with httpPost and httpContext 

       HttpResponse httpResponse = httpClient.execute(httpPost); 

       // checking the CookieStore after logging in 

       List<Cookie> cookies = httpClient.getCookieStore().getCookies(); 

       cook = cookies.toString(); 

       // to store only the value of cookie i.e 26 character long 
       cook = cook.substring(38, 64); 
       Editor editor = sharedPreferences.edit(); 
       editor.putString("cookie", cook); 
       editor.commit(); 

       BufferedReader reader = new BufferedReader(
         new InputStreamReader(httpResponse.getEntity() 
           .getContent(), "UTF-8")); 
       StringBuilder builder = new StringBuilder(); 
       for (String line = null; (line = reader.readLine()) != null;) { 
        builder.append(line + "\n"); 
       } 

       jsonObject = new JSONObject(builder.toString()); 

      } catch (Throwable e) { 
       e.printStackTrace(); 
      } 
      // return result; 
      return jsonObject; 
     } 

     @Override 
     protected void onPostExecute(JSONObject jsonObject) { 
      // TODO Auto-generated method stub 
      super.onPostExecute(jsonObject); 
      pDialog.dismiss(); 

      try { 
       if (jsonObject.getString("firstTime").equalsIgnoreCase("true")) { 

        startActivity(intent); 
       } else { 
        textView.setText("checking for the data"); 
       } 
      } catch (JSONException e) { 
       // TODO Auto-generated catch block 
       e.printStackTrace(); 
      } 

     } 

它背后的原因是什么以及如何在不出错的情况下实现这两个目标？

Answer 1

当您尝试获取并保存cookie并且使jsonObject创建语句不可用并且jsonObject将保持为空时发生异常，因此当您尝试呼叫jsonObject.getString("firstTime")时出现错误。

您可以通过重新安排你的代码解决这个问题，因为遵循

BufferedReader reader = new BufferedReader(
        new InputStreamReader(httpResponse.getEntity() 
          .getContent(), "UTF-8")); 
StringBuilder builder = new StringBuilder(); 
for (String line = null; (line = reader.readLine()) != null;) { 
    builder.append(line + "\n"); 
} 

jsonObject = new JSONObject(builder.toString()); 

// checking the CookieStore after logging in 
List<Cookie> cookies = httpClient.getCookieStore().getCookies(); 
cook = cookies.toString(); 
// to store only the value of cookie i.e 26 character long 
cook = cook.substring(38, 64); 
Editor editor = sharedPreferences.edit(); 
editor.putString("cookie", cook); 
editor.commit(); 

或者只是检查，如果你的cookie不是null试图使用substring方法之前，或者你可以把cookie的事情上尝试， catch块如下

try { 
    // checking the CookieStore after logging in 
    List<Cookie> cookies = httpClient.getCookieStore().getCookies(); 
    cook = cookies.toString(); 
    // to store only the value of cookie i.e 26 character long 
    cook = cook.substring(38, 64); 
    Editor editor = sharedPreferences.edit(); 
    editor.putString("cookie", cook); 
    editor.commit(); 
} catch (Exception e) { 
    Log.e("Cookie Error",e.getMessage()); 
}