Opening app's notification settings in the settings app

I found the answer to this question (albeit helpful) has a bit too much assumed logic. Here is a plain and simple Swift 5 implementation if anyone else stumbles upon this question:

if let bundleIdentifier = Bundle.main.bundleIdentifier, let appSettings = URL(string: UIApplication.openSettingsURLString + bundleIdentifier) {
    if UIApplication.shared.canOpenURL(appSettings) {
        UIApplication.shared.open(appSettings)
    }
}

For Swift 3, use UIApplicationOpenSettingsURLString to go to settings for your app where it shows the Notifications status and "Cellular Data"

let settingsButton = NSLocalizedString("Settings", comment: "")
let cancelButton = NSLocalizedString("Cancel", comment: "")
let message = NSLocalizedString("Your need to give a permission from notification settings.", comment: "")
let goToSettingsAlert = UIAlertController(title: "", message: message, preferredStyle: UIAlertControllerStyle.alert)

goToSettingsAlert.addAction(UIAlertAction(title: settingsButton, style: .destructive, handler: { (action: UIAlertAction) in
    DispatchQueue.main.async {
        guard let settingsUrl = URL(string: UIApplicationOpenSettingsURLString) else {
            return
        }

        if UIApplication.shared.canOpenURL(settingsUrl) {
            if #available(iOS 10.0, *) {
                UIApplication.shared.open(settingsUrl, completionHandler: { (success) in
                    print("Settings opened: \(success)") // Prints true
                })
            } else {
                UIApplication.shared.openURL(settingsUrl as URL)
            } 
        }
    }
}))

logoutUserAlert.addAction(UIAlertAction(title: cancelButton, style: .cancel, handler: nil))

UPDATE: This will be rejected by Apple.

To open notifications part of settings use this

UIApplication.shared.open(URL(string:"App-Prefs:root=NOTIFICATIONS_ID")!, options: [:], completionHandler: nil)