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)
}

只有异常时才会在初始化时崩溃：

指定的网址具有不受支持的方案。仅支持HTTP和HTTPS URL

当我使用url = NSURL(string: "http://www.google.com")时，一切都很好。我实际上正在从API加载URL，因此，我不能确定它们将以http(s)://为前缀。

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

Answer 1

在制作URL的实例之前，请尝试检查SFSafariViewController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 2

在创建url对象之前，您可以在NSUrl字符串中检查 http 的可用性。

将以下代码放在代码之前，它将解决您的问题（您也可以同样方式检查https）

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

Answer 3

我做了Yuvrajsinh＆amp; amp;＆amp;; hoseokchoi的答案。

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（从iOS 11开始），

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

SFSafariViewController崩溃：指定的URL具有不受支持的方案

4 个答案: