how to open an URL in Swift3

IosSwiftSwift3

Ios Problem Overview


openURL has been deprecated in Swift3. Can anyone provide some examples of how the replacement openURL:options:completionHandler: works when trying to open an url?

Ios Solutions


Solution 1 - Ios

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)
}

Solution 2 - Ios

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.

Solution 3 - Ios

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)

Solution 4 - Ios

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)
    }
}

Solution 5 - Ios

import UIKit 
import SafariServices 

let url = URL(string: "https://sprotechs.com")
let vc = SFSafariViewController(url: url!) 
present(vc, animated: true, completion: nil)

Solution 6 - Ios

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.
    }
    
    
};

Solution 7 - Ios

This works fine and doesn't leave the application.

    if let url = URL(string: "https://www.stackoverflow.com") {
    let vc = SFSafariViewController(url: url)
    self.present(vc, animated: true, completion: nil)
}

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionShane O'SeasnainView Question on Stackoverflow
Solution 1 - IosDevran Cosmo UenalView Answer on Stackoverflow
Solution 2 - IosNirav DView Answer on Stackoverflow
Solution 3 - IosChetan RajagiriView Answer on Stackoverflow
Solution 4 - IosSenõr GansoView Answer on Stackoverflow
Solution 5 - IosSalman KhanView Answer on Stackoverflow
Solution 6 - IosScott MaretickView Answer on Stackoverflow
Solution 7 - IosSk.AzadView Answer on Stackoverflow