Swift Open Link in Safari - webview

Question

I am currently opening the link in my app in a WebView, but I'm looking for an option to open the link in Safari instead.

Answer 1

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)

Answer 2

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:

Answer 3

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)
}

Answer 4

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)
}

Answer 5

Swift 3 & IOS 10.2
UIApplication.shared.open(URL(string: "http://www.stackoverflow.com")!, options: [:], completionHandler: nil)
Swift 3 & IOS 10.2

Answer 6

since iOS 10 you should use:
guard let url = URL(string: linkUrlString) else {
return
}
if #available(iOS 10.0, *) {
UIApplication.shared.open(url, options: [:], completionHandler: nil)
} else {
UIApplication.shared.openURL(url)
}

Answer 7

In Swift 1.2:
#IBAction func openLink {
let pth = "http://www.google.com"
if let url = NSURL(string: pth){
UIApplication.sharedApplication().openURL(url)
}

Answer 8

In Swift 2.0:
UIApplication.sharedApplication().openURL(NSURL(string: "http://stackoverflow.com")!)

Answer 9

Swift 5
if let url = URL(string: "https://www.google.com") {
UIApplication.shared.open(url)
}

Answer 10

if your using SwiftUI:
Link("Stack Overflow", destination: URL(string: "https://www.stackoverflow.com/")!)

Answer 11

IOS 11.2 Swift 3.1- 4
let webView = WKWebView()
override func viewDidLoad() {
super.viewDidLoad()
guard let url = URL(string: "https://www.google.com") else { return }
webView.frame = view.bounds
webView.navigationDelegate = self
webView.load(URLRequest(url: url))
webView.autoresizingMask = [.flexibleWidth,.flexibleHeight]
view.addSubview(webView)
}
func webView(_ webView: WKWebView, decidePolicyFor navigationAction: WKNavigationAction, decisionHandler: #escaping (WKNavigationActionPolicy) -> Void) {
if navigationAction.navigationType == .linkActivated {
if let url = navigationAction.request.url,
let host = url.host, !host.hasPrefix("www.google.com"),
UIApplication.shared.canOpenURL(url) {
UIApplication.shared.open(url)
print(url)
print("Redirected to browser. No need to open it locally")
decisionHandler(.cancel)
} else {
print("Open it locally")
decisionHandler(.allow)
}
} else {
print("not a user click")
decisionHandler(.allow)
}
}