如何加载 Web 视图中的 url 链接并将其保留在 SWIFT 中的该 Web 视图中

How to load a url link that is inside a web view and keep it in that web view in SWIFT(如何加载 Web 视图中的 url 链接并将其保留在 SWIFT 中的该 Web 视图中)

问题描述

我有一个网页视图，它的网页底部带有链接.我有办法处理它们被点击，但它所做的只是将它们打开到一个新的 safari 浏览器中.这不是我想要的.我希望他们留在 Web 视图中并允许我像那样轻松地导航它们.有没有办法做到这一点?

I have a web view that has a webpage with links at the bottom. I have a way to deal with them being clicked, but all it does is open them into a new safari browser. This is not what I want.I would like for them to stay inside the web view and allow me to navigate them easily like that. Is there a way to do this?

我尝试过只使用常规的 Web 视图，但底部的链接不起作用.

I have tried just using a regular web view, but then the links at the bottom will not work.

这是我目前所拥有的:

class ViewController: UIViewController, UIWebViewDelegate {


@IBOutlet weak var myWebview: UIWebView!
@IBOutlet weak var returnButton: UIButton!

override func viewDidLoad() {
    super.viewDidLoad()

    prettyReturnButton(returnButton)

    let url = NSURL(string: "mywebpage")

    myWebview.delegate = self

    print("Web page:" , url)
    let requestObj = NSURLRequest(URL: url!);
    myWebview.loadRequest(requestObj);
    myWebview.keyboardDisplayRequiresUserAction = true



}

func webView(webView: UIWebView, shouldStartLoadWithRequest request: NSURLRequest, navigationType: UIWebViewNavigationType) -> Bool {
    switch navigationType {
    case .LinkClicked:
        // Open links in Safari
        //let newRequestObj = NSURLRequest(URL: request.URL!)
       // webView.loadRequest(newRequestObj)
        UIApplication.sharedApplication().openURL(request.URL!)
        return false
    default:
        // Handle other navigation types...
        return true
    }
}

推荐答案

在 webview 委托中，您将希望使用相同的 webview 导航到新的 url.

In the webview delegate you are going to want to navigate to the new url using that same webview.

对您而言，这意味着您应该在您的开关盒内使用loadRequest

For you this means that inside your switch case you should use loadRequest

loadRequest 文档

要解决您的问题，您应该使用 UIApplication.sharedApplication().openURL(request.URL!)

To fix your issue you should remove the line using UIApplication.sharedApplication().openURL(request.URL!)

并将其替换为对 loadRequest

func webView(webView: UIWebView, shouldStartLoadWithRequest request: NSURLRequest, navigationType: UIWebViewNavigationType) -> Bool {
    switch navigationType {

    case .LinkClicked:
        // Open links in Safari

        guard let newRequest = request as? URLRequest else { return false }

        webView.loadRequest(newRequest)
        return false
    default:
        // Handle other navigation types...
        return true
   }
}

希望这会有所帮助！祝你有美好的一天！

Hope this helps! Have a nice day!

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