SwiftUI: use Link with onTapGesture

Question

I want to specify a Link in my SwiftUI view, but I also want to register when/if that Link was actually tapped or not. What's the best way to do this? I tried adding an onTapGesture on the Link (which is a View I believe):

Link("Test", destination: URL(string: "testThreePanel://")!)
                    .onTapGesture {
                        print("Testing button link tapped")
                    }

But the onTapGesture doesn't get invoked on tapping the link.

Is there another way to do this?

score 3 · Accepted Answer · edited Nov 13 '21 at 04:40

3

you could try this, works for me, but only on iOS15 / macOS12:

// iOS15 / macOS12 only 
Link("Test", destination: URL(string: "https://duckduckgo.com")!)
        .environment(\.openURL, OpenURLAction { url in
            print("---> testing link actioned")
            return .systemAction
        })

edited Nov 13 '21 at 04:40

Z S

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answered Nov 12 '21 at 05:46

workingdog support Ukraine

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I tried this and got an error message: "Key path value type 'WritableKeyPath' cannot be converted to contextual type 'KeyPath'" – Z S Nov 12 '21 at 17:36
Is this supposed to only work in iOS 15.0 / macOS 12.0? Because I would need to support iOS14 / macOS11 as well. – Z S Nov 12 '21 at 17:39
yes you are correct, it works on ios 15, macos 12. There was no mention of io14 in the question. I'll delete my answer if it's no use to you. – workingdog support Ukraine Nov 12 '21 at 23:03
Thanks. You can label the original answer with the fact that it's only for iOS15 / macOS12, and I'll mark it as the answer. I don't believe there's another way to do it at this point. – Z S Nov 12 '21 at 23:32
updated my answer. – workingdog support Ukraine Nov 13 '21 at 01:58

score 2 · Answer 2 · answered Aug 09 '22 at 12:17

Just enable firing both embedded and added gestures simultaneously.

struct ContentView: View {
    var body: some View {
        Link("Test", destination: URL(string: "testThreePanel://")!)
            .simultaneousGesture(
                TapGesture()
                    .onEnded { val in
                        print("Testing button link tapped \(val)")
                    }
            )
    }
}

score 1 · Answer 3 · answered Apr 18 '22 at 08:44

For iOS 15 it also works if you remove the 'url' parameter, just in case you don't want to use it.

Link("Test", destination: URL(string: "https://duckduckgo.com")!)
.environment(\.openURL, OpenURLAction { _ in
                print("---> testing link actioned")
                return .systemAction
            })

SwiftUI: use Link with onTapGesture

3 Answers3

Linked