Question

我正在尝试在前景中获取应用程序的名称。我通过运行后台服务来实现这一点，我已经实现了Runnable每秒运行一次这个服务。除了在Jelly Bean之外，这在所有操作系统中都能正常工作。我的服务被杀了。我知道如果一个应用程序消耗更多的RAM，那么它会杀死可用的后台服务来管理所需的空间。例如高分辨率游戏。但我会丢失数据。我实施了Foreground Services。它的工作正常。我的疑问是，这对物质效率或电池消耗有何影响？除了前景之外还有其他方法吗？

这是我前台提供的服务。

@Override
    public int onStartCommand(Intent intent, int flags, int startId) {
        ReceiverRunning = true;
        context = this;

        Intent intent1 = new Intent(this, ComponentSelector.class);  
          PendingIntent pendingIndent = PendingIntent.getActivity(this, 1, intent1, 0);  
          Notification mNotification = new Notification(R.drawable.ic_launcher, "BackgroundApp", System.currentTimeMillis());  
          mNotification.setLatestEventInfo(this, "BatteryUSage", "Click to open to app", pendingIndent);  
          mNotification.flags = mNotification.flags|Notification.FLAG_ONGOING_EVENT;  
            startForeground(1000, mNotification);

        // Start service if not started.
        if (!ForegroundApp.isRunning == true) {
            context.startService(new Intent(context, Brightness.class));
        }

        boolean has_tele = getPackageManager().hasSystemFeature(
                PackageManager.FEATURE_TELEPHONY);
        if (has_tele == true) {
            TelephonyManager teleman = (TelephonyManager) getBaseContext()
                    .getSystemService(Context.TELEPHONY_SERVICE);
            if (teleman != null)
                deviceId = teleman.getDeviceId();
        }

        uId = deviceInfo();

        try {
            NetworkInfo info = (NetworkInfo) ((ConnectivityManager) context
                    .getSystemService(Context.CONNECTIVITY_SERVICE))
                    .getActiveNetworkInfo();
            if (info != null) {
                Log.d("wifiRun", "Network available");
                ConnectivityManager conMan = (ConnectivityManager) context
                        .getSystemService(Context.CONNECTIVITY_SERVICE);
                NetworkInfo.State wifi = null;
                if (conMan.getNetworkInfo(1).isAvailable())
                    wifi = conMan.getNetworkInfo(1).getState();
                if (wifi == NetworkInfo.State.CONNECTED
                        || wifi == NetworkInfo.State.CONNECTING) {
                    wifiCheck = true;
                    context.startService(new Intent(context, WiFi.class));
                    Log.d("wifiRun","wifiCheck: " +wifiCheck);
                } else {
                    Log.d("wifiRun","wifiCheck: " +wifiCheck);
                }
            }
            } catch (Exception e) {
                e.printStackTrace();
            }

        // Register for Screen On and Screen Off.
        IntentFilter intentFilter = new IntentFilter();
        intentFilter.addAction(Intent.ACTION_SCREEN_OFF);
        intentFilter.addAction(Intent.ACTION_SCREEN_ON);
        sReceiver = new ServiceDefinition();
        registerReceiver(sReceiver, intentFilter);
        //return super.onStartCommand(intent, flags, startId);
        return START_STICKY;
    }

Answer 1

我想，别无他法。您也可以使您的服务变得粘性（http://developer.android.com/reference/android/app/Service.html#START_STICKY），但在这种情况下，使服务前景更正确。当然，您的服务会耗尽电池电量，因此请尽量不要在服务中使用“重型”操作。

如果您将提供服务的源代码，我们将检查您是否实现了所有最佳服务。

服务在Jelly Bean中被杀死

1 个答案: