Swift Open Link in Safari

It's not "baked in to Swift", but you can use standard UIKit methods to do it. Take a look at UIApplication's openUrl(_:) (deprecated) and open(_:options:completionHandler:).

Swift 4 + Swift 5 (iOS 10 and above)

guard let url = URL(string: "https://stackoverflow.com") else { return }UIApplication.shared.open(url)

Swift 3 (iOS 9 and below)

guard let url = URL(string: "https://stackoverflow.com") else { return }UIApplication.shared.openURL(url)

Swift 2.2

guard let url = URL(string: "https://stackoverflow.com") else { return }UIApplication.sharedApplication().openURL(url)    

New with iOS 9 and higher you can present the user with a SFSafariViewController (see documentation here). Basically you get all the benefits of sending the user to Safari without making them leave your app. To use the new SFSafariViewController just:

import SafariServices

and somewhere in an event handler present the user with the safari view controller like this:

let svc = SFSafariViewController(url: url)present(svc, animated: true, completion: nil)

The safari view will look something like this:

UPDATED for Swift 4: (credit to Marco Weber)

if let requestUrl = NSURL(string: "http://www.iSecurityPlus.com") {     UIApplication.shared.openURL(requestUrl as URL) }

OR go with more of swift style using guard:

guard let requestUrl = NSURL(string: "http://www.iSecurityPlus.com") else {    return}UIApplication.shared.openURL(requestUrl as URL) 

Swift 3:

You can check NSURL as optional implicitly by:

if let requestUrl = NSURL(string: "http://www.iSecurityPlus.com") {     UIApplication.sharedApplication().openURL(requestUrl)}