Make button open link - Swift - ios

Question

This is the code I have now, taken from an answer to a similar question.
#IBAction func GoogleButton(sender: AnyObject) {
if let url = NSURL(string: "www.google.com"){
UIApplication.sharedApplication().openURL(url)
}
}
The button is called Google Button and its text is www.google.com
How do I make it open the link when I press it?

Answer 1

What your code shows is the action that would occur once the button is tapped, rather than the actual button. You need to connect your button to that action.
(I've renamed the action because GoogleButton is not a good name for an action)
In code:
override func viewDidLoad() {
super.viewDidLoad()
googleButton.addTarget(self, action: "didTapGoogle", forControlEvents: .TouchUpInside)
}
#IBAction func didTapGoogle(sender: AnyObject) {
UIApplication.sharedApplication().openURL(NSURL(string: "http://www.google.com")!)
}
In IB:
Edit: in Swift 3, the code for opening a link in safari has changed. Use UIApplication.shared().openURL(URL(string: "http://www.stackoverflow.com")!) instead.
Edit: in Swift 4
UIApplication.shared.openURL(URL(string: "http://www.stackoverflow.com")!)

Answer 2

The string you are supplying for the NSURL does not include the protocol information. openURL uses the protocol to decide which app to open the URL.
Adding "http://" to your string will allow iOS to open Safari.
#IBAction func GoogleButton(sender: AnyObject) {
if let url = NSURL(string: "http://www.google.com"){
UIApplication.sharedApplication().openURL(url)
}
}

Answer 3

if let url = URL(string: "your URL") {
if #available(iOS 10, *){
UIApplication.shared.open(url)
}else{
UIApplication.shared.openURL(url)
}
}

Answer 4

as openUrl method is deprecated in iOS 10, here is solution for iOS 10
let settingsUrl = NSURL(string:UIApplicationOpenSettingsURLString) as! URL
UIApplication.shared.open(settingsUrl, options: [:], completionHandler: nil)

Answer 5

In Swift 4
if let url = URL(string: "http://yourURL") {
UIApplication.shared.open(url, options: [:])
}

Answer 6

if iOS 9 or higher it's better to use SafariServices, so your user will not leave your app.
import SafariServices
let svc = SFSafariViewController(url: url)
present(svc, animated: true, completion: nil)

Answer 7

For Swift 3.0:
if let url = URL(string: strURlToOpen) {
UIApplication.shared.openURL(url)
}

Answer 8

This code works with Xcode 11
if let url = URL(string: "http://www.google.com") {
UIApplication.shared.open(url, options: [:])
}

Answer 9

The code that you have should open the link just fine. I believe, that you probably just copy-pasted this code fragment into your code. The problem is that the UI component (button) in the interface (in storyboard, most likely) is not connected to the code. So the system doesn't know, that when you press the button, it should call this code.
In order to explain this fact to the system, open the storyboard file, where your Google Button is located, then in assistant editor open the file, where your func GoogleButton code fragment is located. Right-click on the button, and drag the line to the code fragment.
If you create this button programmatically, you should add target for some event, for instance, UITouchUpInside. There are plenty of examples on the web, so it shouldn't be a problem :)
UPDATE: As others noted already, you should also add a protocol to the link ("http://" or "https://"). It will do nothing otherwise.

Answer 10

For Swift3 , below code is working fine
#IBAction func Button(_ sender: Any) {
UIApplication.shared.open(urlStore1, options: [:], completionHandler: nil)
}

Answer 11

Actually You Can Use It Like This In Your Action Button Works For Swift 5 :
guard let settingsUrl = URL(string:"https://yourLink.com") else {
return
}
UIApplication.shared.open(settingsUrl, options: [:], completionHandler: nil)
}

Answer 12

// How to open a URL in Safari
import SafariServices \\ import
#IBAction func google(_ sender: Any)
{
if let url = URL(string: "https://www.google.com")
{
let safariVC = SFSafariViewController(url: url)
present(safariVC, animated: true, completion: nil)
}
}

Answer 13

I made this way:
I imported SafariServices
import SafariServices
First step: I defined a button just above viewDidLoad:
let myButton = UIButton()
Second step: I called a function inside viewDidLoad:
func setupMyButton() {
view.addSubview(myButton)
myButton.configuration = .plain()
myButton.configuration?.cornerStyle = .capsule
myButton.configuration?.title = "Go to Google"
myButton.addTarget(self, action: #selector(selector), for: .touchUpInside)
myButton.translatesAutoresizingMaskIntoConstraints = false
NSLayoutConstraint.activate([
myButton.centerXAnchor.constraint(equalTo: view.centerXAnchor),
myButton.centerYAnchor.constraint(equalTo: view.centerYAnchor),
myButton.widthAnchor.constraint(equalToConstant: 200),
myButton.heightAnchor.constraint(equalToConstant: 50),
])
}
Third step: At the bottom of the scope, I called an #objc func to use as selector. (Outside viewDidLoad)
#objc func selector() {
if let url = URL(string: "https://www.google.com")
{
let safariVC = SFSafariViewController(url: url)
present(safariVC, animated: true, completion: nil)
}
}
And I did not forget to call my func at the beginning of the viewDidLoad:
setupMyButton()
A dude named PRAVEEN BHATI helped me at the third step.
Hope this helps.