在WKWebView中启动电话/电子邮件/地图链接

问题描述:

KINWebBrowser是一款适用于iOS应用的开源Web浏览器模块。我最近升级KINWebBrowser使用WKWebView开始逐步取消UIWebView。这就产生了显著改善,但:在WKWebView中启动电话/电子邮件/地图链接

问题:WKWebView并不能使用户推出包含电话号码,电子邮件地址,地图等的URL链接

如何配置一个WKWebView启动标准iOS行为这些替代网址是否以展示页面的链接形式启动?

所有在WKWebKit

code is available here

更多资讯,请参阅issue on the KINWebBrowser GitHub here

+1

你不能这样做。如果这个功能对你来说很重要,那么这就是现在坚持使用UIWebView的原因 - 以及向Apple提交增强请求的原因。 UIWebView可以做WKWebView无法做到的事情。 – matt 2014-10-22 19:07:04

我能得到它的谷歌地图链接工作(这似乎与目标= “_blank”)和电话:方案通过将此功能添加到您的KINWebBrowserViewController.m

- (void)webView:(WKWebView *)webView decidePolicyForNavigationAction:(WKNavigationAction *)navigationAction decisionHandler:(void (^)(WKNavigationActionPolicy))decisionHandler 
{ 
    if(webView != self.wkWebView) { 
     decisionHandler(WKNavigationActionPolicyAllow); 
     return; 
    } 

    UIApplication *app = [UIApplication sharedApplication]; 
    NSURL   *url = navigationAction.request.URL; 

    if (!navigationAction.targetFrame) { 
     if ([app canOpenURL:url]) { 
      [app openURL:url]; 
      decisionHandler(WKNavigationActionPolicyCancel); 
      return; 
     } 
    } 
    if ([url.scheme isEqualToString:@"tel"]) 
    { 
     if ([app canOpenURL:url]) 
     { 
      [app openURL:url]; 
      decisionHandler(WKNavigationActionPolicyCancel); 
      return; 
     } 
    } 
    decisionHandler(WKNavigationActionPolicyAllow); 
} 
+2

这段代码有点冗长,并不保证会调用'decisionHandler'。 – 2014-10-22 19:16:02

+0

好点!谢谢。这是一个深夜,来自两个不同来源的最后一分钟补丁一起投掷,以快速解决问题。虽然它确实有效,但并不是最干净的,肯定存在decisionHandler没有被调用的情况。我编辑了代码。谢谢! – 2014-10-23 13:00:58

+2

此代码绝对是正确的路径,但它引入了一个安全漏洞,可能会导致不希望发起呼叫和FaceTime呼叫。看看我的解释在这里:https://github.com/dfmuir/KINWebBrowser/issues/10 – dfmuir 2014-10-25 06:15:42

这有助于我的Xcode 8 WKWebview

func webView(_ webView: WKWebView, createWebViewWith configuration: WKWebViewConfiguration, for navigationAction: WKNavigationAction, windowFeatures: WKWindowFeatures) -> WKWebView? { 
    if navigationAction.targetFrame == nil { 
     let url = navigationAction.request.url 
     if url?.description.range(of: "http://") != nil || url?.description.range(of: "https://") != nil || url?.description.range(of: "mailto:") != nil || url?.description.range(of: "tel:") != nil { 
      UIApplication.shared.openURL(url!) 
     } 
    } 
    return nil 
} 

编辑:

在链接必须属性target="_blank"

以上回答workes我,但我需要它改写为SWIFT 2.3

if navigationAction.targetFrame == nil { 
    let url = navigationAction.request.mainDocumentURL 
    if url?.description.rangeOfString("mailto:")?.startIndex != nil || 
     url?.description.rangeOfString("tel:")?.startIndex != nil 
    { 
     if #available(iOS 10, *) { 
      UIApplication.sharedApplication().openURL(url!,options: [:], completionHandler: nil) 
     } else { 
      UIApplication.sharedApplication().openURL(url!) // deprecated 
     } 
    } 
} 

工程上的Xcode 8.1,雨燕2.3。

对于target =“_ blank”,电话号码(tel :)和电子邮件(mailto :)链接。

func webView(webView: WKWebView, decidePolicyForNavigationAction navigationAction: WKNavigationAction, decisionHandler: (WKNavigationActionPolicy) -> Void) { 
    if webView != self.webview { 
     decisionHandler(.Allow) 
     return 
    } 

    let app = UIApplication.sharedApplication() 
    if let url = navigationAction.request.URL { 
     // Handle target="_blank" 
     if navigationAction.targetFrame == nil { 
      if app.canOpenURL(url) { 
       app.openURL(url) 
       decisionHandler(.Cancel) 
       return 
      } 
     } 

     // Handle phone and email links 
     if url.scheme == "tel" || url.scheme == "mailto" { 
      if app.canOpenURL(url) { 
       app.openURL(url) 
       decisionHandler(.Cancel) 
       return 
      } 
     } 

     decisionHandler(.Allow) 
    } 
} 
+0

不要忘记添加委托:WKUIDelegate和var webView = WKWebView()属性 – 2016-11-22 13:03:49

您需要实施的其他回调得到这个权利(雨燕3.1):

// Gets called if webView cant handle URL 
func webView(_ webView: WKWebView, didFailProvisionalNavigation navigation: WKNavigation!, withError error: Error) { 
    guard let failingUrlStr = (error as NSError).userInfo["NSErrorFailingURLStringKey"] as? String else { return } 
    let failingUrl = URL(string: failingUrlStr)! 

    switch failingUrl { 
    // Needed to open Facebook 
    case _ where failingUrlStr.startsWith("fb:"): 
    if #available(iOS 10.0, *) { 
     UIApplication.shared.open(failingUrl, options: [:], completionHandler: nil) 
     return 
    } // Else: Do nothing, iOS 9 and earlier will handle this 

    // Needed to open Mail-app 
    case _ where failingUrlStr.startsWith("mailto:"): 
    if UIApplication.shared.canOpenURL(failingUrl) { 
     UIApplication.shared.openURL(failingUrl) 
     return 
    } 

    // Needed to open Appstore-App 
    case _ where failingUrlStr.startsWith("itmss://itunes.apple.com/"): 
    if UIApplication.shared.canOpenURL(failingUrl) { 
     UIApplication.shared.openURL(failingUrl) 
     return 
    } 

    default: break 
    } 
} 

现在的Facebook,邮件,苹果商店,...得到直接从您的应用程序称为无需打开Safari