Question

我需要创建一个允许我的应用程序工作的服务，当我关闭它时，我已尝试使用STICKY_SERVICE但它不起作用...如果有人可以描述我如何做到这一点请回答这个问题

它适用于Android 7.1，而不适用于其他版本

这是我的代码......

public class SensorService extends Service {
public int counter=0;

public SensorService(Context applicationContext) {
    super();
    Log.i("HERE", "here I am!");
}

public SensorService() {
}

@Override
public int onStartCommand(Intent intent, int flags, int startId) {
    super.onStartCommand(intent, flags, startId);
    startTimer();
    return START_STICKY;
}

@Override
public void onDestroy() {
    super.onDestroy();
    Log.i("EXIT", "ondestroy!");
    stoptimertask();
    Intent broadcastIntent = new Intent("RestartSensor");
    sendBroadcast(broadcastIntent);
}

private Timer timer;
private TimerTask timerTask;
long oldTime=0;
public void startTimer() {
    //set a new Timer
    timer = new Timer();

    //initialize the TimerTask's job
    initializeTimerTask();

    //schedule the timer, to wake up every 1 second
    timer.schedule(timerTask, 1000, 1000); //
}

/**
 * it sets the timer to print the counter every x seconds
 */
public void initializeTimerTask() {
    timerTask = new TimerTask() {
        public void run() {
            Log.i("in timer", "in timer ++++  "+ (counter++));
        }
    };
}

/**
 * not needed
 */
public void stoptimertask() {
    //stop the timer, if it's not already null
    if (timer != null) {
        timer.cancel();
        timer = null;
    }
}

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

这是我的服务Restarter ...

public class SensorRestarterBroadcastReceiver extends BroadcastReceiver {

@Override
public void onReceive(Context context, Intent intent) {
    Log.i(SensorRestarterBroadcastReceiver.class.getSimpleName(), "Service Stops");
    context.startService(new Intent(context, SensorService.class));
}
}

和mainClass

public class MainActivity extends AppCompatActivity {

Intent mServiceIntent;
private SensorService mSensorService;

Context ctx;

public Context getCtx() {
    return ctx;
}


@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    ctx = this;
    setContentView(R.layout.activity_main);
    mSensorService = new SensorService(getCtx());
    mServiceIntent = new Intent(getCtx(), mSensorService.getClass());
    if (!isMyServiceRunning(mSensorService.getClass())) {
        startService(mServiceIntent);
    }
}

private boolean isMyServiceRunning(Class<?> serviceClass) {
    ActivityManager manager = (ActivityManager) getSystemService(Context.ACTIVITY_SERVICE);
    for (ActivityManager.RunningServiceInfo service : manager.getRunningServices(Integer.MAX_VALUE)) {
        if (serviceClass.getName().equals(service.service.getClassName())) {
            Log.i ("isMyServiceRunning?", true+"");
            return true;
        }
    }
    Log.i ("isMyServiceRunning?", false+"");
    return false;
}


@Override
protected void onDestroy() {
    stopService(mServiceIntent);

    Log.i("MAINACT", "onDestroy!");
    super.onDestroy();
}
}

这会正确地重新启动服务，但是在3/4秒后它就会死掉。

我已将此添加到我的清单

<service
        android:name=".SensorService"
        android:enabled="true" >
    </service>

    <receiver
        android:name=".SensorRestarterBroadcastReceiver"
        android:enabled="true"
        android:exported="true"
        android:label="RestartServiceWhenStopped">
        <intent-filter>
            <action android:name="RestartSensor"/>
        </intent-filter>
    </receiver>

Answer 1

将以下代码添加到传感器服务中，并根据需要进行编辑。您需要将粘性服务绑定到通知以使服务保持活动状态并从前台开始。

@覆盖 public void onCreate（）{ super.onCreate（）;

    Intent notifIntent = new Intent(this, SensorService.class);
    PendingIntent pi = PendingIntent.getActivity(this, 0, notifIntent, 0);

    Notification notification = new NotificationCompat.Builder(this)
            .setSmallIcon(R.drawable.smslogo_100x100)
            .setColor(ContextCompat.getColor(this, R.color.colorAccent))
            .setContentTitle(getResources().getString(R.string.app_name))
            .setContentText("Running")
            .setContentIntent(pi)
            .build();
    startForeground(101010, notification);

}

这将纠正您所面临的问题，服务将永远运行。

如何在android中创建后台服务？

1 个答案: