Swift3中已弃用openURL
。谁能提供一些例子,说明当尝试打开url时,替换openURL:options:completionHandler:
是如何工作的?
发布于 2016-09-17 20:36:16
你所需要的是:
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)
}
发布于 2019-04-11 00:45:52
import UIKit
import SafariServices
let url = URL(string: "https://sprotechs.com")
let vc = SFSafariViewController(url: url!)
present(vc, animated: true, completion: nil)
发布于 2017-01-03 09:07:54
我使用的是Swift macOS (v10.12.1) Xcodev8.1 Swift 3.0.1,下面是我在ViewController.swift中使用的方法:
//
// 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.
}
};
https://stackoverflow.com/questions/39546856
复制相似问题