为什么我的服务不开始？

Question

我使用AlarmManager启动服务，但出于某种原因，我无法弄清楚服务未启动。

在我的onResume（）的代码用于启动该服务是：

Intent appIntent = new Intent(cxt, NotificationService.class); 

      AlarmManager alarmManager = (AlarmManager) getSystemService(ALARM_SERVICE); 
      PendingIntent penIntent = PendingIntent.getService(this, 0, 
        appIntent, PendingIntent.FLAG_UPDATE_CURRENT); 

      alarmManager.cancel(penIntent); 

      penIntent = PendingIntent.getService(this, 0, appIntent, PendingIntent.FLAG_UPDATE_CURRENT); 

      Calendar cal = Calendar.getInstance(); 
      cal.set(Calendar.HOUR_OF_DAY, 5); 
      cal.set(Calendar.MINUTE, 00); 
      cal.set(Calendar.SECOND, 00); 

      alarmManager.setRepeating(AlarmManager.ELAPSED_REALTIME_WAKEUP, cal.getTimeInMillis(), 
        10 * 1000, penIntent); 

清单中包含以下内容：

<uses-permission android:name="android.permission.INTERNET" /> 
<uses-permission android:name="android.permission.RECEIVE_BOOT_COMPLETED" /> 
<uses-permission android:name="com.android.alarm.permission.SET_ALARM" /> 
<uses-permission android:name="android.permission.WAKE_LOCK" /> 

<application 
    android:debuggable="true" 
    android:icon="@drawable/icon" 
    android:label="@string/app_name" > 
     <service 
      android:name="de.brillow.mensa.Speiseplan$NotificationService" 
      android:enabled="true" /> 

最后是服务本身是这样的：

public class NotificationService extends Service { 

     private WakeLock mWakeLock; 

     @Override 
     public IBinder onBind(Intent intent) { 
      return null; 
     } 

     private void handleIntent(Intent intent) { 
      // obtain the wake lock 
      PowerManager pm = (PowerManager) getSystemService(POWER_SERVICE); 
      mWakeLock = pm.newWakeLock(PowerManager.PARTIAL_WAKE_LOCK, 
        "wakelocktag"); 
      mWakeLock.acquire(); 

      Log.i("PowerManager", "acquired"); 

      // do the actual work, in a separate thread 
      new PollTask().execute(); 
     } 

     private class PollTask extends AsyncTask<Void, Void, Void> { 

      @Override 
      protected Void doInBackground(Void... params) { 
       // do stuff! 
       Log.i("Background", "stuff"); 
       return null; 
      } 


      @Override 
      protected void onPostExecute(Void result) { 
       // handle your data 
       int id = 001; 
       Log.i("Noti", "fication"); 
       NotificationManager mNotifyMng = (NotificationManager) getSystemService(NOTIFICATION_SERVICE); 
       NotificationCompat.Builder mBuilder = new NotificationCompat.Builder(
         Speiseplan.this).setSmallIcon(R.drawable.icon) 
         .setContentTitle("Test").setContentText("Test!"); 

       mNotifyMng.notify(id, mBuilder.build()); 
       Log.i("Notification", "sent"); 
       stopSelf(); 
       Log.i("Service", "stopped"); 
      } 
     } 


     @Override 
     public void onStart(Intent intent, int startId) { 
      handleIntent(intent); 
     } 


     @Override 
     public int onStartCommand(Intent intent, int flags, int startId) { 
      handleIntent(intent); 
      return START_NOT_STICKY; 
     } 


     public void onDestroy() { 
      super.onDestroy(); 
      mWakeLock.release(); 
     } 
    } 

由于某种原因，通知和Log.i消息都不出现。我希望你能帮忙。

在此先感谢！

Answer 1

您不应在Service中使用AsyncTask。 AsyncTask只应用于将结果返回到当前活动中的任务，即它应该在Activity或Fragment中定义。这就是它的设计目的。

对于您的情况，您可以将doInBackground(...)和onPostExecute(...)之间的所有内容放入您的服务的onCreate(...)。如果你想使用线程。如果您需要在Activity中显示服务结果，请调用Intent并将任何详细信息转换为附加内容，例如，

调用服务...

Intent intent=new Intent(context,VisualizingActivity.class); 
intent.putExtra("A", "Data A"); 
intent.putExtra("B", "Data B"); 
context.startActivity(intent); 

处理的活动......

String A= intent.getStringExtra("A"); 
String B= intent.getStringExtra("B"); 

如果你只是想创建Notification是你可以从服务直接做到。不需要AsyncTask或Activity。

希望这会有所帮助...干杯！