Adding a action to a UIAlert that takes the user to settings - ios

Question

I have a UIAlert that notifies the user that they do not have an internet connection and that they need one in order to use the app. As well as letting them dismiss the alert by tapping the ok action I also want to have a action that when tapped takes the user to the settings app.
func displayAlert(title: String, message: String){
var formEmpty = UIAlertController(title: title, message: message, preferredStyle: UIAlertControllerStyle.Alert)
formEmpty.addAction((UIAlertAction(title: "Ok", style: .Default, handler: { (action) -> Void in
})))

Answer 1

Use this code . May be help it.
override func viewDidAppear(animated: Bool) {
var alertController = UIAlertController (title: "Title", message: "Go to Settings?", preferredStyle: .Alert)
var settingsAction = UIAlertAction(title: "Settings", style: .Default) { (_) -> Void in
let settingsUrl = NSURL(string: UIApplicationOpenSettingsURLString)
if let url = settingsUrl {
UIApplication.sharedApplication().openURL(url)
}
}
var cancelAction = UIAlertAction(title: "Cancel", style: .Default, handler: nil)
alertController.addAction(settingsAction)
alertController.addAction(cancelAction)
presentViewController(alertController, animated: true, completion: nil);
}
Please note UIApplicationOpenSettingsURLString is only available on iOS8.0 and after so if your app should support iOS7 you'll have to check for availability of the constant (or if using Swift 2.0 use the #availability keyword).

Answer 2

You can navigate to the setting with this code:
let settingsUrl = NSURL(string: UIApplicationOpenSettingsURLString)
UIApplication.sharedApplication().openURL(settingsUrl!)
After adding this code in your function your function will look like:
func displayAlert(title: String, message: String){
var formEmpty = UIAlertController(title: title, message: message, preferredStyle: UIAlertControllerStyle.Alert)
formEmpty.addAction((UIAlertAction(title: "Ok", style: .Default, handler: { (action) -> Void in
//This will call when you press ok in your alertview
let settingsUrl = NSURL(string: UIApplicationOpenSettingsURLString)
UIApplication.sharedApplication().openURL(settingsUrl!)
})))
}

Answer 3

For iOS 10, Swift 3:
let alert = UIAlertController(title: "Alert!", message: "your message here", preferredStyle: .alert)
let settingsAction = UIAlertAction(title: "Settings", style: .default) { (_) -> Void in
let settingsUrl = NSURL(string: UIApplicationOpenSettingsURLString)
UIApplication.shared.open(settingsUrl as! URL, options: [:], completionHandler: nil)
alert.addAction(settingsAction)
present(alert, animated: true, completion: nil)