如何在Swift3中打开一个URL

Question

openURL在Swift3中已被弃用。任何人都可以提供一些替代openURL:options:completionHandler:试图打开一个网址时如何工作的例子吗？

Answer 1

所有你需要的是：

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) 
} 

Answer 2

以上的答案是正确的，但如果你想查询你canOpenUrl还是不要尝试这样的。

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)") 
    }) 
} 

注：如果您不希望处理完成后，您还可以这样写。

UIApplication.shared.open(url, options: [:]) 

无需编写completionHandler因为它包含的是默认值nil，检查apple documentation更多细节。

Answer 3

斯威夫特3版本

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) 
    } 
} 

Answer 4

我使用MacOS的塞拉利昂（v10.12.1）的Xcode V8.1斯威夫特3.0.1，这里是我在ViewController.swift什么工作：

// 
// ViewController.swift 
// UIWebViewExample 
// 
// Created by Scott Maretick on 1/2/17. 
// Copyright © 2017 Scott Maretick. All rights reserved. 
// 

import UIKit 

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. 
    } 


}; 

Answer 5

如果你想打开应用程序本身，而不是离开应用程序，你可以导入SafariServices并解决。

import UIKit 
import SafariServices 
let url = URL(string: "https://www.google.com") 
     let vc = SFSafariViewController(url: url!) 
     present(vc, animated: true, completion: nil)