为什么UIWebView无法在Safari中打开我的链接？

Question

我写了这个代码：

if navigationType == UIWebViewNavigationType.LinkClicked { 
     UIApplication.sharedApplication().openURL(request.URL!) 
     return false 
    } 

，它适用于所有正常的URL。但是当我用下面的代码点击图片/ gif时：

<a href="https://cdn-images-1.medium.com/max/800/1*IR7oti7mEsYunzp_HhcNig.gif" 
target="_blank" class="c" 
style="background-image: url('//cdn.iframe.ly/pj?url=https%3A%2F%2Fcdn-images-1.medium.com%2Fmax%2F800%2F1*IR7oti7mEsYunzp_HhcNig.gif&amp;width=1280&amp; 
key=9b0e65e32a5412cf736c430972d914c33f46250b&amp;cache=temp');"> 
1*IR7oti7mEsYunzp_HhcNig.gif </a> 

它在我的应用程序中，在UIWebView中打开这个gif/URL。

我该如何强制它会在Safari中打开该gif/URL？

Answer 1

必须在代码的其他地方打破某些东西。 做工精细是这样的：

import UIKit 

class ViewController: UIViewController, UIWebViewDelegate { 

@IBOutlet var webView : UIWebView? 

override func viewDidLoad() { 
    super.viewDidLoad() 
    // Do any additional setup after loading the view, typically from a nib. 

    self.webView?.loadHTMLString("<a href='https://cdn-images-1.medium.com/max/800/1*IR7oti7mEsYunzp_HhcNig.gif' target='_blank' class='c' style='background-image: url('//cdn.iframe.ly/pj?url=https%3A%2F%2Fcdn-images-1.medium.com%2Fmax%2F800%2F1*IR7oti7mEsYunzp_HhcNig.gif&amp;width=1280&amp; key=9b0e65e32a5412cf736c430972d914c33f46250b&amp;cache=temp');'> 1*IR7oti7mEsYunzp_HhcNig.gif </a>", baseURL: NSURL(string: "testurl.com")) 
} 

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

     } 

    return true 
} 
}