如何在 android 中创建后台服务？

Question

我需要创建一个服务，让我的应用程序在关闭时也能正常工作，我试过 STICKY_SERVICE 但它不起作用...如果有人能描述我如何做请回答这个问题。

它适用于 android 7.1，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;
}
}

这是我的服务重启器...

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));
}
}

和主类

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();
}
}

这会正确重新启动服务，但在 3/4 秒后它就会死掉。

我已将其添加到我的清单中

<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

将以下代码添加到您的传感器服务并根据您的需要进行编辑。您需要将粘性服务绑定到通知以保持服务处于活动状态并在前台启动。

@Override 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);

}

这将解决您面临的问题，服务将运行永远。

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