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How to customize the way links are opened

Paul Hudson @twostraws December 1st 2022

Updated for Xcode 14.2

When your user taps a URL shown inside a SwiftUI Text or Link view, it will open in Safari by default. However, you can customize this behavior by replacing the openURL environment key – you might want to handle the link entirely, or perhaps pass it back to the system to open once your custom action completes.

For example, this code adjusts both a Link and a Text view so that all URLs are sent to a handleURL() method to be acted on:

struct ContentView: View {
    var body: some View {
        VStack {
            Link("Visit Apple", destination: URL(string: "https://apple.com")!)
            Text("[Visit Apple](https://apple.com)")
        }
        .environment(\.openURL, OpenURLAction(handler: handleURL))
    }

    func handleURL(_ url: URL) -> OpenURLAction.Result {
        print("Handle \(url) somehow")
        return .handled
    }
}

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As you can see, handleURL() returns a OpenURLAction.Result value of .handled, which means the method accepted the link and acted on it. There are alternatives:

Use .discarded if you mean you weren’t able to handle the link.
Use .systemAction if you want to trigger the default behavior, perhaps in addition to your own logic.
Use .systemAction(someOtherURL) if you want to open a different URL using the default behavior, perhaps a modified version of the URL that was originally triggered.

Remember, links will use your app’s accent color by default, but you can change that using the tint() modifier:

Text("[Visit Apple](https://apple.com)")
    .tint(.indigo)

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How to customize the way links are opened

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