SFSafariViewController crash: The specified URL has an unsupported scheme

Question

My code:

if let url = NSURL(string: \"www.google.com\") {
    let safariViewController = SFSafariViewController(URL: url)
    safariViewController.view.tintColor         


        

Answer 1

I did a combination of Yuvrajsinh's & hoseokchoi's answers.

func openLinkInSafari(withURLString link: String) {

    guard var url = NSURL(string: link) else {
        print("INVALID URL")
        return
    }

    /// Test for valid scheme & append "http" if needed
    if !(["http", "https"].contains(url.scheme.lowercaseString)) {
        let appendedLink = "http://".stringByAppendingString(link)

        url = NSURL(string: appendedLink)!
    }

    let safariViewController = SFSafariViewController(URL: url)
    presentViewController(safariViewController, animated: true, completion: nil)
}

Answer 2

Use WKWebView's method (starting iOS 11),

class func handlesURLScheme(_ urlScheme: String) -> Bool

Answer 3

Try checking scheme of URL before making an instance of SFSafariViewController.

Swift 3:

func openURL(_ urlString: String) {
    guard let url = URL(string: urlString) else {
        // not a valid URL
        return
    }

    if ["http", "https"].contains(url.scheme?.lowercased() ?? "") {
        // Can open with SFSafariViewController
        let safariViewController = SFSafariViewController(url: url)
        self.present(safariViewController, animated: true, completion: nil)
    } else {
        // Scheme is not supported or no scheme is given, use openURL
        UIApplication.shared.open(url, options: [:], completionHandler: nil)
    }
}

Swift 2:

func openURL(urlString: String) {
    guard let url = NSURL(string: urlString) else {
        // not a valid URL
        return
    }

    if ["http", "https"].contains(url.scheme.lowercaseString) {
        // Can open with SFSafariViewController
        let safariViewController = SFSafariViewController(URL: url)
        presentViewController(safariViewController, animated: true, completion: nil)
    } else {
        // Scheme is not supported or no scheme is given, use openURL
        UIApplication.sharedApplication().openURL(url)
    }
}

Answer 4

You can check for availability of http in your url string before creating NSUrl object.

Put following code before your code and it will solve your problem (you can check for https also in same way)

var strUrl : String = "www.google.com"
if strUrl.lowercaseString.hasPrefix("http://")==false{
     strUrl = "http://".stringByAppendingString(strUrl)
}