Webview - Open IOS App with Google Chrome

Question

Hello, developing an app for IOS, but would like it to open with Google Chrome instead of Safari browser.

Could one of you help me?

Code:

import UIKit
import WebKit

class ViewController: UIViewController, WKNavigationDelegate {

    var webView: WKWebView!

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

    override func viewDidLoad() {
        super.viewDidLoad()

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


    } 

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

        //Aqui você tem a URL, e pode fazer o que quiser com ela.

        decisionHandler(.allow)
    }
}

Answer 1

Replace the last function with this one, it will check if it is possible to open with Chrome and send the link

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

    guard let requestURL = navigationAction.request.url else {
        decisionHandler(.allow)
        return
    }

    if(navigationAction.navigationType ==  WKNavigationType.linkActivated) {
        // Remove http:// ou https:// e substitui com googlechrome://
        let newLinkWithHttp = requestURL.absoluteString.replacingOccurrences(of: "http://", with: "googlechrome://")
        let newLinkWithHttps = newLinkWithHttp.replacingOccurrences(of: "https://", with: "googlechrome://")

        // Verifica se é possível abrir com o chrome, caso não seja abre o link normal com o safari
        if UIApplication.shared.canOpenURL(URL(string: newLinkWithHttps)!) {
            UIApplication.shared.open(URL(string: newLinkWithHttps)!, options: [:], completionHandler: nil)
        } else {
            UIApplication.shared.open(requestURL, options: [:], completionHandler: nil)
        }

        decisionHandler(.cancel)
        return
    }

    decisionHandler(.allow)
}