Swift Open Link in Safari
Swift 5
Swift 5: Check using canOpneURL
if valid then it's open.
guard let url = URL(string: "https://iosdevcenters.blogspot.com/") else {
return
}
if UIApplication.shared.canOpenURL(url) {
UIApplication.shared.open(url, options: [:], completionHandler: nil)
}
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)
}
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:
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)