All you need is:
guard let url = URL(string: "http://www.google.com") else {
return //be safe
}
if #available(iOS 10.0, *) {
UIApplication.shared.open(url, options: [:], completionHandler: nil)
} else {
UIApplication.shared.openURL(url)
}
I'm using macOS Sierra (v10.12.1) Xcode v8.1 Swift 3.0.1 and here's what worked for me in ViewController.swift:
//
// ViewController.swift
// UIWebViewExample
//
// Created by Scott Maretick on 1/2/17.
// Copyright © 2017 Scott Maretick. All rights reserved.
//
import UIKit
import WebKit
class ViewController: UIViewController {
//added this code
@IBOutlet weak var webView: UIWebView!
override func viewDidLoad() {
super.viewDidLoad()
// Your webView code goes here
let url = URL(string: "https://www.google.com")
if UIApplication.shared.canOpenURL(url!) {
UIApplication.shared.open(url!, options: [:], completionHandler: nil)
//If you want handle the completion block than
UIApplication.shared.open(url!, options: [:], completionHandler: { (success) in
print("Open url : \(success)")
})
}
}
override func didReceiveMemoryWarning() {
super.didReceiveMemoryWarning()
// Dispose of any resources that can be recreated.
}
};
import UIKit
import SafariServices
let url = URL(string: "https://sprotechs.com")
let vc = SFSafariViewController(url: url!)
present(vc, animated: true, completion: nil)
Swift 3 version
import UIKit
protocol PhoneCalling {
func call(phoneNumber: String)
}
extension PhoneCalling {
func call(phoneNumber: String) {
let cleanNumber = phoneNumber.replacingOccurrences(of: " ", with: "").replacingOccurrences(of: "-", with: "")
guard let number = URL(string: "telprompt://" + cleanNumber) else { return }
UIApplication.shared.open(number, options: [:], completionHandler: nil)
}
}
If you want to open inside the app itself instead of leaving the app you can import SafariServices and work it out.
import UIKit
import SafariServices
let url = URL(string: "https://www.google.com")
let vc = SFSafariViewController(url: url!)
present(vc, animated: true, completion: nil)
Above answer is correct but if you want to check you canOpenUrl
or not try like this.
let url = URL(string: "http://www.facebook.com")!
if UIApplication.shared.canOpenURL(url) {
UIApplication.shared.open(url, options: [:], completionHandler: nil)
//If you want handle the completion block than
UIApplication.shared.open(url, options: [:], completionHandler: { (success) in
print("Open url : \(success)")
})
}
Note: If you do not want to handle completion you can also write like this.
UIApplication.shared.open(url, options: [:])
No need to write completionHandler
as it contains default value nil
, check apple documentation for more detail.
Source: Stackoverflow.com