Question

我目前正在WebView打开我的应用中的链接，但我正在寻找一个选项来打开 Safari 中的链接。

Answer 1

它没有＆＃34;烘焙到Swift＆＃34;，但您可以使用标准的UIKit方法来完成它。看看UIApplication's openUrl()。

Swift 4

guard let url = URL(string: "https://stackoverflow.com") else { return }
UIApplication.shared.open(url)

Swift 3

guard let url = URL(string: "https://stackoverflow.com") else { return }
UIApplication.shared.openURL(url)

Swift 2.2

guard let url = URL(string: "https://stackoverflow.com") else { return }
UIApplication.sharedApplication().openURL(url)

Answer 2

iOS 9及更高版本的新功能，您可以向用户显示SFSafariViewController（请参阅文档here）。基本上，您可以获得将用户发送到Safari而不会让他们离开您的应用程序的所有好处。要使用新的SFSafariViewController：

import SafariServices

以及事件处理程序中的某个位置向用户显示safari视图控制器，如下所示：

let svc = SFSafariViewController(url: url)
present(svc, animated: true, completion: nil)

safari视图将如下所示：

Answer 3

更新了Swift 4 :(归功于Marco Weber）

if let requestUrl = NSURL(string: "http://www.iSecurityPlus.com") {
     UIApplication.shared.openURL(requestUrl as URL) 
}

或者使用guard更多的快捷风格：

guard let requestUrl = NSURL(string: "http://www.iSecurityPlus.com") else {
    return
}

UIApplication.shared.openURL(requestUrl as URL)

斯威夫特3：

您可以通过以下方式隐式检查NSURL是否为可选项：

if let requestUrl = NSURL(string: "http://www.iSecurityPlus.com") {
     UIApplication.sharedApplication().openURL(requestUrl)
}

Answer 4

Swift 3＆amp; IOS 10.2

UIApplication.shared.open(URL(string: "http://www.stackoverflow.com")!, options: [:], completionHandler: nil)

斯威夫特3＆amp; IOS 10.2

Answer 5

因为iOS 10你应该使用：

guard let url = URL(string: linkUrlString) else {
        return
    }

    if #available(iOS 10.0, *) {
        UIApplication.shared.open(url, options: [:], completionHandler: nil)
    } else {
        UIApplication.shared.openURL(url)
    }

Answer 6

快速5

Swift 5: Check using canOpneURL if valid then it's open.

guard let url = URL(string: "https://iosdevcenters.blogspot.com/") else {
     return
 }
if UIApplication.shared.canOpenURL(url) {
     UIApplication.shared.open(url, options: [:], completionHandler: nil)
 }

Answer 7

在Swift 1.2中：

{{1}}

Answer 8

在Swift 2.0中：

UIApplication.sharedApplication().openURL(NSURL(string: "http://stackoverflow.com")!)

Answer 9

雨燕5

if let url = URL(string: "https://www.google.com") {
    UIApplication.shared.open(url)
}

Answer 10

IOS 11.2 Swift 3.1-4

let webView = WKWebView()
override func viewDidLoad() {
    super.viewDidLoad()
    guard let url = URL(string: "https://www.google.com") else { return }
    webView.frame = view.bounds
    webView.navigationDelegate = self
    webView.load(URLRequest(url: url))
    webView.autoresizingMask = [.flexibleWidth,.flexibleHeight]
    view.addSubview(webView)
}
func webView(_ webView: WKWebView, decidePolicyFor navigationAction: WKNavigationAction, decisionHandler: @escaping (WKNavigationActionPolicy) -> Void) {
    if navigationAction.navigationType == .linkActivated  {
        if let url = navigationAction.request.url,
            let host = url.host, !host.hasPrefix("www.google.com"),
            UIApplication.shared.canOpenURL(url) {
            UIApplication.shared.open(url)
            print(url)
            print("Redirected to browser. No need to open it locally")
            decisionHandler(.cancel)
        } else {
            print("Open it locally")
            decisionHandler(.allow)
        }
    } else {
        print("not a user click")
        decisionHandler(.allow)
    }
}

}

Safari中的Swift Open Link

10 个答案: