Android开机自启动服务的实现方法

时间:2022-02-08 08:57:34

  android实现开机自启动可能是移动操作系统中最简单的了,我们只需要监听一个开机启动的Broadcast(广播)即可。首先写一个Receiver(即广播监听器),继承BroadcastReceiver,如下所示:

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复制代码 代码如下:


   public class BootReceiver extends BroadcastReceiver {

 

   private PendingIntent mAlarmSender;

   @Override

   public void onReceive(Context context, Intent intent) {

   // 在这里干你想干的事(启动一个Service,Activity等),本例是启动一个定时调度程序,每30分钟启动一个Service去更新数据

   mAlarmSender = PendingIntent.getService(context, 0, new Intent(context,

   RefreshDataService.class), 0);

   long firstTime = SystemClock.elapsedRealtime();

   AlarmManager am = (AlarmManager) context

   getSystemService(Activity.ALARM_SERVICE);

   am.cancel(mAlarmSender);

   am.setRepeating(AlarmManager.ELAPSED_REALTIME_WAKEUP, firstTime,

   30 * 60 * 1000, mAlarmSender);

   }

   }


接下来,我们只需要在应用程序配置文件AndroidManifest.xml中注册这个Receiver来监听系统启动事件即可,如下所示:

 

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复制代码 代码如下:


   < receiver Android:name=".service.BootReceiver">

 

   < intent-filter>

   < !-- 系统启动完成后会调用-->

   < action android:name="Android.intent.action.BOOT_COMPLETED">

   < /action>

   < /intent-filter>

   < /receiver>