6

I am trying to close SFSafariViewController when I reach a certain page.

But I am unable to do it until the "Done" button is pressed explicitly by the user.

What I want is to get the URL as soon as it reached a certain page and then dismiss the view controller. Then I can pick the rest using this 

func safariViewControllerDidFinish(_ controller: SFSafariViewController){
    // do work here
}

Note: This question has been asked before but I was unable to find any satisfactory answers

Moaz Khan
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  • Have you seen any functionality like this in any other app? I doubt that it's possible. – ullstrm Nov 13 '17 at 08:15
  • I have seen it in some custom APIs where they use callback url which calls this function in app delegate func application(_ app: UIApplication, open url: URL, options: [UIApplicationOpenURLOptionsKey : Any] = [:]) -> Bool {} and where they call dismiss – Moaz Khan Nov 13 '17 at 08:20
  • check this question https://stackoverflow.com/questions/41349184/how-to-close-sfsafariviewcontroller-automatically-when-reaching-a-certain-page – Moaz Khan Nov 13 '17 at 08:23
  • Then you have it there :) I think that's the only way of doing it – ullstrm Nov 13 '17 at 08:34
  • Please can any one tell anything about this? How to get certain URL in SFSafariViewController? – Anil shukla Apr 27 '18 at 14:00
  • @Anilshukla you can't. You need to implement deep linking. – Moaz Khan Apr 27 '18 at 14:01
  • can you give me something to analysis with? That would be helpful ! – Anil shukla Apr 27 '18 at 17:21

3 Answers3

4

If you register a URI scheme for your app, you could have have a link on a page which takes you back to your app. At that point you'll be in

func application(_ app: UIApplication, open url: URL, options: [UIApplicationOpenURLOptionsKey : Any] = [:]) -> Bool {

of your AppDelegate and you can do whatever you wish.

“Because since Safari View Controller has access to a user’s credentials, synced across all of their devices with iCloud Keychain, logging in is going to be a breeze.[…] It takes two steps. The first is where you would’ve used your own in-app browser, just present an instance of SFSafariViewController.

And once the user is finished logging in and the third-party web service redirects back to your app with the custom URL scheme that you fed it, you can accept that in your AppDelegate‘s handleOpenURL method.

From there you can inspect the response and dismiss the instance of SFSafariViewController because you know that the authentication is done.

That’s it. Two steps.”

Source: http://asciiwwdc.com/2015/sessions/504#t=1558.026

Please note the handleOpenURL method is deprecated in favor of application:openURL:options:.

DBD
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    this method is not called – Vyachaslav Gerchicov May 31 '18 at 13:53
  • @Vyachaslav - I added more information. Why do you say the method is not called? – DBD Jun 08 '18 at 15:07
  • the problem is for example in my case is instagram api allows to use http/https URL scheme. In the same time it seems this method is called for custom scheme but not for http/https. And there is almost no info about it in public sources. – Vyachaslav Gerchicov Jun 11 '18 at 13:24
  • Right. That would be a problem. You need to use a custom scheme to get back to the app, like `myapp://`. You could still accomplish this if you had a web site. Have the instagram api send the user to your web site/page which has an unload redirect to the custom scheme used by your app. It's an extra bounce, but this would effectively take you back to your app. – DBD Jun 11 '18 at 13:32
  • the problem is instagram redirects to this website with a token in URL. So I can't redirect it to my app directly - it seems I should encode-decode it manually (because `app:openURL:...` methods work with URL only not with request body – Vyachaslav Gerchicov Jun 11 '18 at 13:51
4

EDIT:

The answer became invalid ... to do so nowadays you have to implement wkwebview component

====================================

Maybe It's late But I think It will help anyone search for the same question Listening to the SFSafariViewControllerDelegate delegate you can use this method inside

func safariViewController(_ controller: SFSafariViewController, initialLoadDidRedirectTo URL: URL) {}

inside it you can make if condition on your certain page you need to take action for ... as example if I want to close it when reaching certain page I'will fo this

 func safariViewController(_ controller: SFSafariViewController, initialLoadDidRedirectTo URL: URL) {
    if URL.absoluteString == "WHATEVER YOUR LINK IS"{
        controller.dismiss(animated: true, completion: nil)
    }
}
abdallah Nader
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2

You cannot do that with a SFSafariViewController. Use a WKWebView instead and implement the WKNavigationDelegate method like optional func webView(_ webView: WKWebView, didFinish navigation: WKNavigation!) and then read the current URL. Define your action based on that.

You can use this to close the view controller that houses your web view. You will unfortunately have to build your own navigation buttons (the ones the SFSafariViewController has), if you want to allow the user access to those. You don't have to provide them.

erik_m_martens
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