Swift Open Link in Safari

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

User · Accepted Answer

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

User · Answer

Swift 5 Swift 5  Check using canOpneURL if valid then it s open  guard let url   URL string   quot https   iosdevcenters blogspot com  quot   else        return    if UIApplication shared canOpenURL url         UIApplication shared open url  options       completionHandler  nil

User · Answer

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

User · Answer

In Swift 2 0   UIApplication sharedApplication   openURL NSURL string   http   stackoverflow com

User · Answer

Swift 3  amp  IOS 10 2  UIApplication shared open URL string   http   www stackoverflow com     options       completionHandler  nil       Swift 3  amp  IOS 10 2

User · Answer

In Swift 1 2    IBAction func openLink           let pth    http   www google com      if let url   NSURL string  pth           UIApplication sharedApplication   openURL url

User · Answer

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

User · Answer

IOS 11 2 Swift 3 1- 4 let webView   WKWebView    override func viewDidLoad         super viewDidLoad       guard let url   URL string   quot https   www google com quot   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  - gt  Void        if navigationAction navigationType     linkActivated            if let url   navigationAction request url              let host   url host   host hasPrefix  quot www google com quot                UIApplication shared canOpenURL url                UIApplication shared open url              print url              print  quot Redirected to browser  No need to open it locally quot               decisionHandler  cancel            else               print  quot Open it locally quot               decisionHandler  allow                  else           print  quot not a user click quot           decisionHandler  allow

User · Answer

Swift 5 if let url   URL string   quot https   www google com quot         UIApplication shared open url

User · Answer

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

[webview] Swift Open Link in Safari

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