SwiftUI: How do I make a button open a URL in safari?

Question

I want a button inside my app, that will open safari and a specific URL address. How do I do this? UIApplication.sharedApplication doesn't seem to work in SwiftUI.

This is my first question here. Please give feedback if I did something wrong

Answer 1

iOS 14.0+

Using Link:

Link("Some label", destination: URL(string: "https://www.mylink.com")!)

Older versions

Button(action: {
    if let url = URL(string: "https://www.mylink.com") {
       UIApplication.shared.open(url)
    }
}) {
    Text("Some label")
}

Answer 2

In the View:

@Environment(\.openURL) private var openURL

Later, e.g. on button press:

guard let url = URL(string: urlString) else { return }
openURL(url)

Read more: https://developer.apple.com/documentation/swiftui/openurlaction

Answer 3

If you are using a SceneDelegate you can handle URL addresses using the following:

func scene(_ scene: UIScene, openURLContexts URLContexts: Set<UIOpenURLContext>) {

    guard let url = URLContexts.first?.url else {
        return
    }

    //Handle URL here
}

From there you can call openURL like you normally would:

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

Answer 4

This can be done through SwiftUI's Link: https://developer.apple.com/documentation/swiftui/link.

Answer 5

In SwiftUI, you can use Link. For example:

Link("Link Text", destination: URL(string: "https://www.example.com/page.html")!)