SFSafariViewController 崩溃：指定的 URL 有一个不受支持的方案

Question

我的代码：

if let url = NSURL(string: "www.google.com") {
    let safariViewController = SFSafariViewController(URL: url)
    safariViewController.view.tintColor = UIColor.primaryOrangeColor()
    presentViewController(safariViewController, animated: true, completion: nil)
}

这仅在初始化时崩溃，但有以下例外：

The specified URL has an unsupported scheme. Only HTTP and HTTPS URLs are supported

当我使用url = NSURL(string: "http://www.google.com")时，一切都很好。我实际上是从 API 加载 URL，因此，我不能确定它们是否会以 http(s)://.

为前缀

如何解决这个问题？我应该始终检查并添加前缀 http://，还是有解决方法？

Answer 1

您可以在创建 NSUrl 对象之前检查 url 字符串中 http 的可用性。

将下面的代码放在你的代码之前，它会解决你的问题（你也可以用同样的方式检查https）

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

Answer 2

在创建 SFSafariViewController 的实例之前尝试检查 URL 的方案。

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 3

我结合了 Yuvrajsinh 和 hoseokchoi 的回答。

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 4

使用WKWebView的方法（开始iOS11），

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

Answer 5

你可以添加到

  NSString* webStringURL = [url stringByAddingPercentEscapesUsingEncoding: NSUTF8StringEncoding];
  NSURL *URL = [NSURL URLWithString: webStringURL];

SFSafariViewController 崩溃：指定的 URL 有一个不受支持的方案

SFSafariViewController crash: The specified URL has an unsupported scheme

ios

swift

sfsafariviewcontroller