Android:如何在收到新通知时打开屏幕？

Question

我已经尝试了几乎所有的东西,我无法理解在后台模式下收到新通知时如何打开手机屏幕.

我正在使用FCM.当应用程序处于后台时,收到的通知由系统托盘管理.当收到新内容时,如何让系统托盘打开屏幕？谢谢!

Answer 1

最后,答案就在这里.

如果您使用Firebase API将通知从服务器发送到设备,请记住,如果您发送如下通知,扩展FirebaseMessagingService中的onMessageReceived()方法仅在BACKGROUND中调用:

                        $msg = array
                        (
                            'body'  => $newsObj['description'],
                        );
                        $fields = array
                        (
                            'registration_ids'  => $registrationIds,
                            'data'          => $msg,
                            'priority'  => "high"
                        );

您会看到,如果将" data "属性更改为" notification ",则仅当应用程序位于FOREGROUND时才会调用onMessageReceived()方法.

要在收到新通知时打开屏幕,请将以下代码放入onMessageReceived()方法:

        // Turn on the screen for notification
        PowerManager powerManager = (PowerManager) getSystemService(POWER_SERVICE);
        boolean result= Build.VERSION.SDK_INT>= Build.VERSION_CODES.KITKAT_WATCH&&powerManager.isInteractive()|| Build.VERSION.SDK_INT< Build.VERSION_CODES.KITKAT_WATCH&&powerManager.isScreenOn();

        if (!result){
            PowerManager.WakeLock wl = powerManager.newWakeLock(PowerManager.FULL_WAKE_LOCK |PowerManager.ACQUIRE_CAUSES_WAKEUP |PowerManager.ON_AFTER_RELEASE,"MH24_SCREENLOCK");
            wl.acquire(10000);
            PowerManager.WakeLock wl_cpu = powerManager.newWakeLock(PowerManager.PARTIAL_WAKE_LOCK,"MH24_SCREENLOCK");
            wl_cpu.acquire(10000);
        }

在AndroidManifest.xml中:

    <uses-permission android:name="android.permission.WAKE_LOCK" />

就这样!