问题
My code:
if let url = NSURL(string: "www.google.com") {
let safariViewController = SFSafariViewController(URL: url)
safariViewController.view.tintColor = UIColor.primaryOrangeColor()
presentViewController(safariViewController, animated: true, completion: nil)
}
This crashes on initialization only with exception:
The specified URL has an unsupported scheme. Only HTTP and HTTPS URLs are supported
When I use url = NSURL(string: "http://www.google.com")
, everything is fine.
I am actually loading URL's from API and hence, I can't be sure that they will be prefixed with http(s)://
.
How to tackle this problem? Should I check and prefix http://
always, or there's a workaround?
回答1:
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)
}
回答2:
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)
}
}
回答3:
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)
}
回答4:
Use WKWebView's method (starting iOS 11),
class func handlesURLScheme(_ urlScheme: String) -> Bool
来源:https://stackoverflow.com/questions/32864287/sfsafariviewcontroller-crash-the-specified-url-has-an-unsupported-scheme