ios - WKWebView abre enlaces de cierto dominio en safari

swift

Question

Buscando ayuda con mi aplicación iOS. Dentro de mi aplicación, quiero abrir enlaces cliqueados desde mi dominio (EX: communionchapelefca.org) en WKWebView y luego tener enlaces de todos los demás dominios (EX: google.com) abiertos en Safari. Preferiría hacer esto también de manera progrógica ya que así es como mi código ya está configurado.

He encontrado algunas soluciones en stackoverflow ( here , here , here y here ), pero todas parecen estar basadas en Obj-C y estoy buscando una solución con Swift. gracias por adelantado.

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

Actualización de Swift 4 para la respuesta de George Vardikos:

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 2

Aquí hay un código de muestra de la respuesta al swift escrito en obj c.

- (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 3

Haga una función para decidir dónde cargar la 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)
    }
}

Uso:

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

Answer 4

Mi solución swift 3:

    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)
        }

    }

No olvide también configurar el delegado

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

Answer 5

Para 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 6

Puede implementar WKNavigationDelegate , agregar el método decidePolicyForNavigationAction y verificar allí el tipo de navegación y la URL solicitada. He usado google.com a continuación, pero puedes cambiarlo a tu dominio:

Xcode 8.3 • Swift 3.1 o posterior

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 7

Xcode 9.3, Swift 4

La fuente: 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)
}}