如何創建Android應用程序，以響應在睡眠模式下手機的物理按鈕的點擊

Question

我想創建一個應用程序，當用戶在手機處於睡眠模式時在主屏幕按鈕中點擊3次以上時，該應用程序將被激活。我如何在Android中實現這一點？

Answer 1

您需要創建一個後臺服務。請注意，您還需要創建一個通知該服務時，這將是可見的運行：

public class MyApp extends Application { 
    @Override 
    public void onCreate() { 
     startService(new Intent(this, BgService.class)); 
    } 
} 

：

public class BgService extends Service { 
     @Override 
     public int onStartCommand(Intent i, int flags, int startId) { 
      startForeground(C.MAIN_SERVICE_NOTIFICATION_ID, buildNotification(getString(R.string.service_title))); 
      return START_NOT_STICKY; 
     } 

     protected Notification buildNotification(String content) { 
      NotificationCompat.Builder builder = new NotificationCompat.Builder(this); 
      builder.setTicker(getString(R.string.app_name)) 
        .setContentTitle(getString(R.string.app_name)) 
        .setContentText(content) 
        .setSmallIcon(R.drawable.notification_icon) 
        .setLargeIcon(BitmapFactory.decodeResource(getResources(), R.drawable.app_icon)) 
        .setWhen(System.currentTimeMillis()) 
        .setAutoCancel(false) 
        .setOngoing(true) 
        .setContentIntent(pendingIntent); 

      if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.JELLY_BEAN) 
       builder.setPriority(Notification.PRIORITY_HIGH); 

      Notification notification = builder.build(); 
      notification.flags |= Notification.FLAG_NO_CLEAR; 
      return notification; 
     } 
    } 

您可以從您的應用程序啓動此服務