Question

寻求iOS应用的一些帮助。在我的应用程序中，我想在WKWebView中打开我的域内的点击链接（EX：communionchapelefca.org），然后在Safari中打开所有其他域（EX：google.com）的链接。我更喜欢以程序方式执行此操作，因为这是我的代码已经设置的方式。

我在stackoverflow上找到了一些解决方案（here，here，here和here），但它们似乎都是基于Obj-C的，我是使用Swift寻找解决方案。提前谢谢。

ViewController.swift：

import UIKit
import WebKit

class ViewController: UIViewController {

    override func viewDidLoad() {
        super.viewDidLoad()

        let myWebView:WKWebView = WKWebView(frame: CGRectMake(0, 0,   UIScreen.mainScreen().bounds.width, UIScreen.mainScreen().bounds.height))

        myWebView.loadRequest(NSURLRequest(URL: NSURL(string: "http://www.communionchapelefca.org/app-home")!))

        self.view.addSubview(myWebView)

Answer 1

您可以实施WKNavigationDelegate，添加decidePolicyForNavigationAction方法并检查navigationType和请求的网址。我在下方使用了 google.com ，但您可以将其更改为您的域名：

Xcode 8.3•Swift 3.1或更高版本

import UIKit
import WebKit

class ViewController: UIViewController, WKNavigationDelegate {

    let webView = WKWebView()

    override func viewDidLoad() {
        super.viewDidLoad()

        webView.frame = view.bounds
        webView.navigationDelegate = self

        let url = URL(string: "https://www.google.com")!
        let urlRequest = URLRequest(url: url)

        webView.load(urlRequest)
        webView.autoresizingMask = [.flexibleWidth,.flexibleHeight]
        view.addSubview(webView)
    }

    func webView(_ webView: WKWebView, decidePolicyFor navigationAction: WKNavigationAction, decisionHandler: @escaping (WKNavigationActionPolicy) -> Void) {
        if navigationAction.navigationType == .linkActivated  {
            if let url = navigationAction.request.url,
                let host = url.host, !host.hasPrefix("www.google.com"),
                UIApplication.shared.canOpenURL(url) {
                UIApplication.shared.open(url)
                print(url)
                print("Redirected to browser. No need to open it locally")
                decisionHandler(.cancel)
            } else {
                print("Open it locally")
                decisionHandler(.allow)
            }
        } else {
            print("not a user click")
            decisionHandler(.allow)
        }
    }
}

Answer 2

对于Swift 3.0

import UIKit
import WebKit

class ViewController: UIViewController, WKNavigationDelegate {
    let wv = WKWebView(frame: UIScreen.main.bounds)
    override func viewDidLoad() {
        super.viewDidLoad()
        guard let url =  NSURL(string: "https://www.google.com") else { return }
        wv.navigationDelegate = self
        wv.load(NSURLRequest(url: url as URL) as URLRequest)
        view.addSubview(wv)
    }

    override func didReceiveMemoryWarning() {
        super.didReceiveMemoryWarning()
    }

    func webView(_ webView: WKWebView, decidePolicyFor navigationAction: WKNavigationAction, decisionHandler: @escaping (WKNavigationActionPolicy) -> Void) {
        if navigationAction.navigationType == .LinkActivated  {
            if let newURL = navigationAction.request.url,
                let host = newURL.host , !host.hasPrefix("www.google.com") &&
                UIApplication.shared.canOpenURL(newURL) &&
                UIApplication.shared.openURL(newURL) {
                    print(newURL)
                    print("Redirected to browser. No need to open it locally")
                    decisionHandler(.cancel)
            } else {
                print("Open it locally")
                decisionHandler(.allow)
            }
        } else {
            print("not a user click")
            decisionHandler(.allow)
        }
    }
}

Answer 3

以下是来自obj c中写的swift响应的示例代码。

- (void)webView:(WKWebView *)webView decidePolicyForNavigationAction:(nonnull WKNavigationAction *)navigationAction decisionHandler:(nonnull void (^)(WKNavigationActionPolicy))decisionHandler
{
    if (navigationAction.navigationType == WKNavigationTypeLinkActivated) {
        if (navigationAction.request.URL) {
            NSLog(@"%@", navigationAction.request.URL.host);
            if (![navigationAction.request.URL.resourceSpecifier containsString:@"ex path"]) { 
                if ([[UIApplication sharedApplication] canOpenURL:navigationAction.request.URL]) {
                    [[UIApplication sharedApplication] openURL:navigationAction.request.URL];
                    decisionHandler(WKNavigationActionPolicyCancel);
                }
            } else {
                decisionHandler(WKNavigationActionPolicyAllow);
            }
        }
    } else {
        decisionHandler(WKNavigationActionPolicyAllow);
    }
}

Answer 4

我的快速3解决方案：

    public func webView(_ webView: WKWebView, decidePolicyFor navigationAction: WKNavigationAction, decisionHandler: @escaping (WKNavigationActionPolicy) -> Void) {

        let url = navigationAction.request.url

        if url?.description.lowercased().range(of: "http://") != nil || url?.description.lowercased().range(of: "https://") != nil {
            decisionHandler(.cancel)
            UIApplication.shared.openURL(url!)
        } else {
            decisionHandler(.allow)
        }

    }

不要忘记设置te委托

    public override func loadView() {
        let webConfiguration = WKWebViewConfiguration()
        webView = WKWebView(frame: .zero, configuration: webConfiguration)
        webView.uiDelegate = self
        webView.navigationDelegate = self
        view = webView
    }

Answer 5

创建一个函数来决定加载URL的位置：

func loadURLString(str: String) {
    guard let url = NSURL(string: str) else {
        return
    }

    if url.host == "www.communionchapelefca.org" {
        // Open in myWebView
        myWebView.loadRequest(NSURLRequest(URL: url))
    } else {
        // Open in Safari
        UIApplication.sharedApplication().openURL(url)
    }
}

用法：

loadURLString("http://www.communionchapelefca.org/app-home") // Open in myWebView
loadURLString("http://www.apple.com") // Open in Safari

Answer 6

George Vardikos答案的Swift 4更新：

public func webView(_ webView: WKWebView, decidePolicyFor navigationAction: WKNavigationAction, decisionHandler: @escaping (WKNavigationActionPolicy) -> Void) {
        let url = navigationAction.request.url
        guard url != nil else {
            decisionHandler(.allow)
            return
        }

        if url!.description.lowercased().starts(with: "http://") ||
            url!.description.lowercased().starts(with: "https://")  {
            decisionHandler(.cancel)
            UIApplication.shared.open(url!, options: [:], completionHandler: nil)
        } else {
            decisionHandler(.allow)
        }
}

Answer 7

我的 2 美分用于 swift 5 和 iOS >10 （ UIApplication.shared.openURL(url!) 已弃用）

 public func webView(_ webView: WKWebView, decidePolicyFor navigationAction: WKNavigationAction, decisionHandler: @escaping (WKNavigationActionPolicy) -> Void) {
        
        guard let url = navigationAction.request.url else{
            decisionHandler(.allow)
            return
        }
        
        let urlString = url.absoluteString.lowercased()
        if urlString.starts(with: "http://") || urlString.starts(with: "https://") {
            decisionHandler(.cancel)
            UIApplication.shared.open(url, options: [:])
        } else {
            decisionHandler(.allow)
        }
        
    }

Answer 8

Xcode 9.3，Swift 4

来源： https://developer.apple.com/documentation/webkit/wkwebview

import UIKit
import WebKit

class ViewController: UIViewController, WKUIDelegate {

var webView: WKWebView!

override func loadView() {
    let webConfiguration = WKWebViewConfiguration()
    webView = WKWebView(frame: .zero, configuration: webConfiguration)
    webView.uiDelegate = self
    view = webView
}

override func viewDidLoad() {
    super.viewDidLoad()

    let myURL = URL(string: "https://www.apple.com")
    let myRequest = URLRequest(url: myURL!)
    webView.load(myRequest)
}}

WKWebView在safari中打开来自某个域的链接

8 个答案: