如何创建Android应用程序，以响应在睡眠模式下手机的物理按钮的点击

Question

我想创建一个应用程序，当用户在手机处于睡眠模式时在主屏幕按钮中点击3次以上时，该应用程序将被激活。我如何在Android中实现这一点？

Answer 1

您需要创建一个后台服务。请注意，您还需要创建一个通知该服务时，这将是可见的运行：

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; 
     } 
    } 

您可以从您的应用程序启动此服务