我已经构建了一个Swift
iOS应用程序,其唯一目的是通过WKWebView来显示网站。WKWebView链接不工作
这工作正常,但按链接(例如在[mailto:]
按钮)不起作用,因为它不能打开任何东西!
有没有人有解决方案?
我已经读了很多关于解决这个问题,但我不知道从哪里开始。
感谢您的帮助
[UPDATE:]
下面的代码示出了溶液
import UIKit
import WebKit
class ViewController: UIViewController, WKNavigationDelegate {
@IBOutlet var containerView : UIView! = nil
var webView: WKWebView?
override func loadView() {
super.loadView()
self.webView = WKWebView()
self.view = self.webView!
}
override func viewDidLoad() {
super.viewDidLoad()
webView?.navigationDelegate = self
let url = NSURL(string:"https://google.de")
let req = NSURLRequest (URL: url!)
self.webView!.loadRequest(req)
}
override func didReceiveMemoryWarning() {
super.didReceiveMemoryWarning()
}
override func prefersStatusBarHidden() -> Bool {
return true
}
func webView(webView: WKWebView, decidePolicyForNavigationAction navigationAction: WKNavigationAction, decisionHandler: (WKNavigationActionPolicy) -> Void) {
let url = navigationAction.request.URL?.absoluteString
let url_elements = url!.componentsSeparatedByString(":")
switch url_elements[0] {
case "mailto":
openCustomApp("mailto://", additional_info: url_elements[1])
decisionHandler(.Cancel)
default:
}
decisionHandler(.Allow)
}
func openCustomApp(urlScheme:String, additional_info:String){
if let requestUrl:NSURL = NSURL(string:"\(urlScheme)"+"\(additional_info)") {
let application:UIApplication = UIApplication.sharedApplication()
if application.canOpenURL(requestUrl) {
application.openURL(requestUrl)
}
}
}
}
看起来很好。你调试了DecisionPolicyForNavigationAction吗?例如的网址? 基本上你只需要mailto://[email protected] – OhadM
你刚刚解决了这个问题。下面的答案是正确的做法,上面的代码显示了解决方案。我唯一的问题是,mailto:链接格式不正确!感谢大家 –