Html 从UIWebView打开Safari中的特定链接

我有一个

UIWebView

，它加载一个本地index.html文件

但是，我想在Safari中打开此html文件中的外部链接，而不是在

UIWebView

中打开

从safari中的

ui按钮打开链接非常简单：

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

从外部链接打开Instagram应用程序也很有魅力

<a href="instagram://media?id=434784289393782000_15903882">instagram://media?id=434784289393782000_15903882</a>

这是你可以做到的

func webView(webView: UIWebView, shouldStartLoadWithRequest request: NSURLRequest, navigationType: UIWebViewNavigationType) -> Bool {
        if let url = request.URL where navigationType == UIWebViewNavigationType.LinkClicked {
            UIApplication.sharedApplication().openURL(url)
            return false
        }
        return true
    }

嗯，似乎还是有点不对劲。我已经添加了我的swift代码，包括你的函数。你能帮我看看出了什么问题吗？非常感谢！您是否已指定，
UIWebView
delegate？
import UIKit

class AccessoriesViewController: UIViewController, UIWebViewDelegate {


    @IBOutlet weak var webView:UIWebView!

    override func viewDidLoad() {
        super.viewDidLoad()


        if let url = NSBundle.mainBundle().URLForResource("accessories", withExtension: "html") {
            webView.loadRequest(NSURLRequest(URL: url))
        }



    }
    override func preferredStatusBarStyle() -> UIStatusBarStyle {
        return UIStatusBarStyle.LightContent;
    }

    func webView(webView: UIWebView, shouldStartLoadWithRequest request: NSURLRequest, navigationType: UIWebViewNavigationType) -> Bool {
        if let url = request.URL where navigationType == UIWebViewNavigationType.LinkClicked {
            UIApplication.sharedApplication().openURL(url)
            return false
        }
        return true
    }

    override func didReceiveMemoryWarning() {
        super.didReceiveMemoryWarning()
        // Dispose of any resources that can be recreated.
    }


}

func webView(webView: UIWebView, shouldStartLoadWithRequest request: NSURLRequest, navigationType: UIWebViewNavigationType) -> Bool {
        if let url = request.URL where navigationType == UIWebViewNavigationType.LinkClicked {
            UIApplication.sharedApplication().openURL(url)
            return false
        }
        return true
    }