如何在收到通知时自动在 Android 上添加新的通知渠道？

Question

如果用户单击通知并打开应用程序，我就可以做到这一点。 我正在使用通知发送频道名称和 ID，然后在 onCreate 中使用它

        if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.O) {
            String channelId = getIntent().getExtras().getString("channelId");
            String channelName = getIntent().getExtras().getString("channelName");
            NotificationChannel channel = new NotificationChannel(channelId, channelName,
                    NotificationManager.IMPORTANCE_HIGH
            );
            getSystemService(NotificationManager.class).createNotificationChannel(channel);
        }

，或者如果应用程序在前台，则调用 onMessageReceived，所以我可以这样做：

        String channelName = remoteMessage.getData().get("channelName");
        String channelId = remoteMessage.getData().get("channelId");
        if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.O) {

            NotificationChannel channel = new NotificationChannel(channelId , channelName ,
                    NotificationManager.IMPORTANCE_HIGH
            );
            getSystemService(NotificationManager.class).createNotificationChannel(channel);
        }

但是如果应用程序关闭或在后台并且用户没有单击在通知上，我无法创建新通道，

这是 python 端：

def sendMulticast(notificationTitle, notificationMessage, notificationLink, notificationTokens, imageLink, channelName, channelId):
   if len(notificationTokens)>0:
        try :
            message = messaging.MulticastMessage(
        tokens = list(dict.fromkeys(notificationTokens)),
        data={"link":notificationLink, "channelName":channelName, "channelId":channelId},
        android=messaging.AndroidConfig(
        priority='high',
        
        notification=messaging.AndroidNotification(title=notificationTitle,
            body=notificationMessage,
            image=imageLink,
            color='#ff0000', default_light_settings="true", priority="max", default_vibrate_timings="true", default_sound ="true", channel_id=channelId
        ),
    ),
        )

            response = messaging.send_multicast(message)
            print('Successfully sent message:', response)
        except Exception as e:
            print(str(e))