Opening app's notification settings in the settings app

Question

In the case that a user may accidentally declines to receive notifications and wants to turn notifications later, how can I use an NSURL to open the IOS Settings App to my app's notification page where they can select Allow Notifications?

Answer 1

Updated 8 Dec, 2021:

This method will open Settings > Your App. It will show all available privacy toggles like camera, photos, notifications, cellular data, etc.

After a comment from @Mischa below, tested and updated the answer to this (more succinct):

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

Previous answer:

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)
    }
}

Answer 2

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))

Answer 3

Swift 5:

if let url = URL(string: UIApplication.openSettingsURLString) {
    UIApplication.shared.open(url)
}

Answer 4

This should keep you covered for all iOS versions, including iOS 15 and iOS 16.

extension UIApplication {
  private static let notificationSettingsURLString: String? = {
    if #available(iOS 16, *) {
      return UIApplication.openNotificationSettingsURLString
    }

    if #available(iOS 15.4, *) {
      return UIApplicationOpenNotificationSettingsURLString
    }

    if #available(iOS 8.0, *) {
      // just opens settings
      return UIApplication.openSettingsURLString
    }

    // lol bruh
    return nil
  }()

  private static let appNotificationSettingsURL = URL(
    string: notificationSettingsURLString ?? ""
  )

  func openAppNotificationSettings() -> Bool {
    guard
      let url = UIApplication.appNotificationSettingsURL,
      self.canOpen(url) else { return false }
    return self.open(url)
  }
}

Usage:

Button {
  let opened = UIApplication.shared.openAppNotificationSettings()
  if !opened {
    print("lol fail")
  }
} label: {
  Text("Notifications")
}

Answer 5

Since iOS 15.4 we can deeplink to the notif settings screen directly using UIApplicationOpenNotificationSettingsURLString:

https://developer.apple.com/documentation/uikit/uiapplicationopennotificationsettingsurlstring?language=objc

Answer 6

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)