Question

我想复制Facebook应用程序的行为，当您点击外部链接时，它会在新的浏览器中打开（但它不在Safari中）。

我目前有一个加载我网站的WKWebView。我可以抓住用户点击事件并打开它们，让我们说在Safari中。我现在想要，而不是在Safari中打开它们，创建一个新的WKWebView并在那里显示链接......

import UIKit
import Darwin
import SafariServices
import WebKit

class ViewController: UIViewController, UIWebViewDelegate,    SFSafariViewControllerDelegate, WKNavigationDelegate {

let url = "http://myurl:3000"
var webView : WKWebView!
var webView2 : WKWebView!

@IBOutlet weak var progressView: UIProgressView!
var myTimer = Timer()
var theBool: Bool = false

override func viewDidLoad() {
    super.viewDidLoad()
    let url = URL(string: self.url)!
    webView.load(URLRequest(url: url) as URLRequest)
    webView.allowsBackForwardNavigationGestures = true
}

override func loadView() {
    webView = WKWebView()
    webView.navigationDelegate = self
    view = webView
}

func webViewDidFinishLoad(_ webView: UIWebView) {
    theBool = true;
}

func webViewDidStartLoad(_ webView: UIWebView) {
    myTimer = Timer.scheduledTimer(timeInterval: 0.06, target: self, selector: #selector(ViewController.updateTimer), userInfo: nil, repeats: true)
    theBool = false;
    progressView.progress = 0.0
}

func webView(_ webView: WKWebView, decidePolicyFor navigationAction: WKNavigationAction, decisionHandler: @escaping (WKNavigationActionPolicy) -> Void) {
    if navigationAction.navigationType == .linkActivated  {
            webView2 = WKWebView()
            webView2.navigationDelegate = self
            view = webView2
            let url = navigationAction.request.url
            webView2.load(URLRequest(url: url!) as URLRequest)

    } else {
        print("not a user click")
        decisionHandler(.allow)
    }
}

func updateTimer()
{
    if (theBool) {
        if (progressView.progress >= 1) {
            progressView.isHidden = true;
            myTimer.invalidate()
        }
        else {
            progressView.progress += 0.1;
        }
    }
    else {

        if (progressView.isHidden){
            progressView.isHidden = false
        }
        progressView.progress += 0.05;
        if (progressView.progress >= 0.60) {
            progressView.progress = 0.60;
        }
    }
}

}

问题是，即使它正在构建，当我点击外部链接时，应用程序崩溃。

libc++abi.dylib: terminating with uncaught exception of type NSException (lldb)

Answer 1

我建议您实现此 WKUIDelegate 方法，并在此方法中的新WKWebView中加载URL。

optional func webView(_ webView: WKWebView, 
createWebViewWith configuration: WKWebViewConfiguration, 
              for navigationAction: WKNavigationAction, 
   windowFeatures: WKWindowFeatures) -> WKWebView?

从WkWebView打开外部链接到新的WKWebview

1 个答案: