从UIWebView打开Safari中的特定链接 [英] Open specific link in Safari from UIWebView

问题描述

我有一个加载本地index.html文件的 UIWebView 。

然而，我有一个外部链接在这个html文件中，我想在Safari中打开而不是在 UIWebView 中内部打开。

打开Safari中的链接来自 UIButton 很简单：

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

<

 < a href =instagram：// p>从外部链接中打开Instagram应用程序也很有用。 ？介质ID = 434784289393782000_15903882\" >的Instagram：//介质ID = 434784289393782000_15903882< / A>？ 
  

所以我的第一个尝试是这样做的：

 < a href =safari：//stackoverflow.com>在Safari中打开< / a> 
  

然而，这似乎并不奏效，然后我读了一些关于使用 webView：shouldStartLoadWithRequest：navigationType：

但是每个设法在Safari中打开外部链接的人都是用Obj-C编写的，太熟悉了，因为我正在写Swift。

使用Swift代码进行更新：

 导入UIKit 
 
类附件视图控制器：UIViewController，UIWebViewDelegate {
 
 
 @IBOutlet弱var webView：UIWebView！ 
 
重写func viewDidLoad（）{
 super.viewDidLoad（）
 
 $ b $如果让url = NSBundle.mainBundle（）。URLForResource（accessories ，withExtension：html）{
 webView.loadRequest（NSURLRequest（URL：url））
} 
 
 
 $ b} 
覆盖func preferredStatusBarStyle（） - > UIStatusBarStyle {
返回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 
} 
返回true 
} 
 
覆盖func didReceiveMemoryWarning（）{
 super.didReceiveMemoryWarning（）
 //处理任何可以重新创建的资源。 
} 
 
 
} 
  

解决方案

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

I have a UIWebView which loads a local index.html file.

However I have an external link in this html file that I'd like to open in Safari instead of internally in the UIWebView.

Opening a link in safari from say a UIButton was simple enough:

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

Opening the Instagram app from a external link also works like a charm.

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

So my first though was to do something like this:

<a href="safari://stackoverflow.com">Open in Safari</a>

However that doesn't seem to work, then I read something about using webView:shouldStartLoadWithRequest:navigationType:

But everyone who's managed to open an external link in Safari is writing in Obj-C which I'm not too familiar with as I'm writing in Swift.

Update with Swift Code:

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.
    }


}

解决方案

Here is how you can do it!

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
    }

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