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How to open any URL clicked within my iOS app in the in-app browser?

How can I set a rule such that when my user clicks any web link within my app, it opens in the in-app browser instead of Safari?

Context: I'm building an app that has several places where links are either embedded by me or are loaded through various user interactions, e.g. a link leading to a page that houses another link. I want to ensure that the user seldom leaves my app and hence want to open all external web links in an in-app browser that has already been developed

Target Build: iOS 11. Environment/ Language: Swift 4

like image 796
zaam Avatar asked Oct 16 '18 03:10

zaam


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2 Answers

Simple solution using SFSafariViewController which available in a separate package

import SafariServices

Swift 4, iOS 9.0+

let url = URL(string: "your URL here")
let vc = SFSafariViewController(url: url)
present(vc, animated: true)

If you need more control - use Configuration parameter (Swift 4, iOS 11.0+):

Example:

let config = SFSafariViewController.Configuration()
config.entersReaderIfAvailable = true

let url = URL(string: "your URL here")
let vc = SFSafariViewController(url: url, configuration: config)
present(vc, animated: true)
like image 139
Petr Lazarev Avatar answered Oct 22 '22 20:10

Petr Lazarev


...it opens in the in-app browser instead of Safari?

In that case, you would need your own WebView. Here's a quick snippet of a simple webView inside a controller:

import UIKit
import WebKit

/// The controller for handling webviews.
class WebViewController: UIViewController {

    // MARK: - Properties
    internal lazy var button_Close: UIButton = {
        let button = UIButton(type: .custom)
        button.setImage(.close, for: .normal)
        button.imageEdgeInsets = UIEdgeInsets.init(top: 0, left: -30, bottom: 0, right: 0)
        button.addTarget(self, action: #selector(back(_:)), for: .touchUpInside)
        return button
    }()

    public var urlString: String! {
        didSet {
            if let url = URL(string: urlString) {
                let urlRequest = URLRequest(url:url)
                self.webView.load(urlRequest)
            }
        }
    }

    private lazy var webView: WKWebView = {
        let webView = WKWebView()
        webView.navigationDelegate = self
        return webView
    }()

    // MARK: - Functions
    // MARK: Overrides

    override func viewDidLoad() {
        super.viewDidLoad()

        let barButton = UIBarButtonItem(customView: self.button_Close)
        self.button_Close.frame = CGRect(x: 0, y: 0, width: 55.0, height: 44.0)
        let negativeSpacer = UIBarButtonItem(barButtonSystemItem: UIBarButtonItem.SystemItem.fixedSpace, target: nil, action: nil)
        if #available(iOS 11.0, *) {
            negativeSpacer.width = -30
        }
        self.navigationItem.leftBarButtonItems = [negativeSpacer, barButton]

        self.view.addSubview(self.webView)
        self.webView.snp.makeConstraints {
            $0.edges.equalToSuperview()
        }
    }

    @objc func back(_ sender: Any) {
        self.dismiss()
    }
}

extension WebViewController: WKNavigationDelegate {
    func webView(_ webView: WKWebView, didStartProvisionalNavigation navigation: WKNavigation!) {
        // Show here a HUD or any loader
    }

    func webView(_ webView: WKWebView, didFinish navigation: WKNavigation!) {
        // Dismiss your HUD
    }
}

and presenting such webViewController, like so (passing a URLString):

let webViewVC = WebViewController()
webViewVC.urlString = "https://www.glennvon.com/"
let navCon = UINavigationController(rootViewController: webViewVC)
self.navigationController?.present(navCon, animated: true, completion: nil)

If you're using storyboard, then simply drag a webview in your controller and setup the delegate. This should help you out :)

like image 43
Glenn Posadas Avatar answered Oct 22 '22 20:10

Glenn Posadas