WKWebView 在 safari 中打开来自特定域的链接

Question

在我的应用程序中，我想在 WKWebView 中打开来自 my 域（例如：communionchapelefca.org）的链接，然后有来自 other 的链接 个域在 Safari 中打开。我更愿意以编程方式执行此操作。

我找到了一些关于 Stack overflow (here, here, here, and here) 的解决方案，但它们似乎都是基于 Obj-C 的，我正在寻找使用 Swift.

的解决方案

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

制作一个函数来决定在哪里加载 URL:

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 2

您可以实施 WKNavigationDelegate，添加 decidePolicyForNavigationAction 方法并检查导航类型并请求 url。我在下面使用了 google.com，但您可以将其更改为您的域：

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)
                return
            } else {
                print("Open it locally")
                decisionHandler(.allow)
                return
            }
        } else {
            print("not a user click")
            decisionHandler(.allow)
            return
        }
    }
}

Answer 3

对于Swift3.0

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 4

这是对用 obj c 编写的 swift 的响应的示例代码。

- (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 5

我的swift3 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)
        }

    }

别忘了设置委托

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

Answer 6

Swift 4 George Vardikos 的回答更新：

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

Swift 5 和 iOS >10

不要使用 UIApplication.shared.openURL(url!), it's deprecated. We use UIApplication.shared.open(url!)。

 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)
        }
        
    }

WKWebView 在 safari 中打开来自特定域的链接

WKWebView open links from certain domain in safari

safari

ios

swift

wkwebview

Swift 5 和 iOS >10