Android时AsyncTask如果没有互联网发现过程如何给出警告“请连接互联网”

Question

Android AsyncTask处理，发现没有互联网连接系统发出警告“用户请连接互联网”。我如何在我的代码中执行此操作？

另一个问题是，当互联网没有连接其显示加载，但我不回页面。如何解决这个问题呢？

我不明白如何在此代码中执行此操作。

public class data extends AppCompatActivity { 

    private String TAG = MainActivity.class.getSimpleName(); 

    private ProgressDialog pDialog; 
    private ListView lv; 

    // URL to get contacts JSON 
    private static String url = "http://orangehostbd.com/app/bpl2/index.php"; 

    ArrayList<HashMap<String, String>> contactList; 

    @Override 
    protected void onCreate(Bundle savedInstanceState) { 
     super.onCreate(savedInstanceState); 
     setContentView(R.layout.activity_data); 

     contactList = new ArrayList<>(); 

     lv = (ListView) findViewById(R.id.list); 

     new GetContacts().execute(); 
    } 

    /** 
    * Async task class to get json by making HTTP call 
    */ 
    private class GetContacts extends AsyncTask<Void, Void, Void> { 

     @Override 
     protected void onPreExecute() { 
      super.onPreExecute(); 
      // Showing progress dialog 
      pDialog = new ProgressDialog(data.this); 
      pDialog.setMessage("Please wait..."); 
      pDialog.setCancelable(false); 
      pDialog.show(); 

     } 



     @Override 
     protected Void doInBackground(Void... arg0) { 
      HttpHandler sh = new HttpHandler(); 

      // Making a request to url and getting response 
      String jsonStr = sh.makeServiceCall(url); 

      Log.e(TAG, "Response from url: " + jsonStr); 

      if (jsonStr != null) { 
       try { 
        JSONObject jsonObj = new JSONObject(jsonStr); 

        // Getting JSON Array node 
        JSONArray contacts = jsonObj.getJSONArray("contacts"); 

        // looping through All Contacts 
        for (int i = 0; i < contacts.length(); i++) { 
         JSONObject c = contacts.getJSONObject(i); 

         String id = c.getString("id"); 
         String name = c.getString("name"); 
         String email = c.getString("email"); 
         String address = c.getString("address"); 
         String gender = c.getString("gender"); 

         // Phone node is JSON Object 
         JSONObject phone = c.getJSONObject("phone"); 
         String mobile = phone.getString("mobile"); 
         String home = phone.getString("home"); 
         String office = phone.getString("office"); 

         // tmp hash map for single contact 
         HashMap<String, String> contact = new HashMap<>(); 

         // adding each child node to HashMap key => value 
         contact.put("id", id); 
         contact.put("name", name); 
         contact.put("email", email); 
         contact.put("mobile", mobile); 

         // adding contact to contact list 
         contactList.add(contact); 
        } 
       } catch (final JSONException e) { 
        Log.e(TAG, "Json parsing error: " + e.getMessage()); 
        runOnUiThread(new Runnable() { 
         @Override 
         public void run() { 
          Toast.makeText(getApplicationContext(), 
            "Json parsing error: " + e.getMessage(), 
            Toast.LENGTH_LONG) 
            .show(); 
         } 
        }); 

       } 
      } else { 
       Log.e(TAG, "Couldn't get json from server."); 
       runOnUiThread(new Runnable() { 
        @Override 
        public void run() { 
         Toast.makeText(getApplicationContext(), 
           "Couldn't get json from server. Check LogCat for possible errors!", 
           Toast.LENGTH_LONG) 
           .show(); 
        } 
       }); 

      } 

      return null; 
     } 

     @Override 
     protected void onPostExecute(Void result) { 
      super.onPostExecute(result); 
      // Dismiss the progress dialog 
      if (pDialog.isShowing()) 
       pDialog.dismiss(); 
      /** 
      * Updating parsed JSON data into ListView 
      * */ 
      ListAdapter adapter = new SimpleAdapter(
        data.this, contactList, 
        R.layout.list_item, new String[]{"name", "email", 
        "mobile"}, new int[]{R.id.name, 
        R.id.email, R.id.mobile}); 

      lv.setAdapter(adapter); 
     } 

    } 
} 

Answer 1

1. 创建Helper类。 在这个类中，创建一个方法来检查Internet连接。

   public static boolean isNetworkConnected(Context context) { 
       final ConnectivityManager connectivityManager = ((ConnectivityManager) context.getSystemService(Context.CONNECTIVITY_SERVICE)); 
       return connectivityManager.getActiveNetworkInfo() != null && connectivityManager.getActiveNetworkInfo().isConnected(); 
   } 
   

2. 内，您的活动类，使用上面的doInBackgroundisNetworkConnected()方法。

   if (Helper.isNetworkConnected(this)) { 
       // do your stuff 
   } else{ 
       Toast.makeText(YourActivity.this, "Your internet connection is lost.", Toast.LENGTH_SHORT).show(); 
   } 
   

3. 在Manifest文件中使用此权限。

   <uses-permission android:name="android.permission.INTERNET" /> 
   

Answer 2

如果问题是“如何显示内部doInBackground警告”，因为你不能访问GUI内doInBackground技术上可以使用progress indicator of asynctask到邮件传输到GUI。

但是我会recommond一个differen形式给出：启动异步任务

要么

• check for problems之前还是让异步任务停止并经@Muthukrishnan拉金德伦

描述返回错误。