How to correctly open URL from UIAlertViewController's handler?

Question

I'm trying to open URL after clicking on alert's button. I've tried different approaches, but none of it worked. Here's my code:

let alert = UIAlertController(title: "Warning", message: "Do you want to open the link?", preferredStyle: .Alert);

alert.addAction(UIAlertAction(title: "Cancel", style: .Default, handler: nil));
let okAction = UIAlertAction(title: "Open in Safari", style: .Default) {(action) in
     UIApplication.sharedApplication().openURL(NSURL(string: "www.google.com")!);
     return;
});
alert.addAction(okAction);

self.presentViewController(alert, animated: true, completion: nil);

If I put some simple function like println, it works fine.

Answer 1

For Swift 3 Users, this is how to have a button that open a link in an alert:

let action: UIAlertAction = UIAlertAction(title: "SomeTitle", style: .default, handler: {
   (action) in
      UIApplication.shared.open(URL(string: "http://www.google.com")!, options: [:], completionHandler: nil)
      NSLog("Opening link")
 })

Answer 2

I was able to get mine to work by creating a variable called "link" with a string of the url:

let link = 'http://google.com"

and i set my addAction to look like:

showAlert.addAction(UIAlertAction(title: "Download", 
   style: UIAlertActionStyle.Default, handler: { 
       (action:UIAlertAction!) -> Void in 
            UIApplication.sharedApplication().openURL(NSURL(string: link)!)
                  print("something here... button click or action logging")
 }))

hope this helps someone... took me forever to figure outg how tgo get the handler to work right...