I'm trying to use:
UIApplication.shared.open(URL(string:"App-Prefs:root=NOTIFICATIONS_ID")!, options: [:], completionHandler: nil)
But it's just sending me to the Settings app, not to the Notification tab.
I have tried this without luck.
Any solution?
Apple will Reject your app submission. Because this is the private api. Use the below function in the below. ref: How do I open phone settings when a button is clicked?
guard let settingsUrl = URL(string: UIApplication.openSettingsURLString) else {
return
}
if UIApplication.shared.canOpenURL(settingsUrl) {
UIApplication.shared.open(settingsUrl, completionHandler: { (success) in
print("Settings opened: \(success)") // Prints true
})
}