SwiftUI: popover to persist (not be dismissed when tapped outside)

Question

I created this popover:

import SwiftUI

struct Popover : View {
  @State var showingPopover = false

  var body: some View {

    Button(action: {
      self.showingPopover = true
    }) {
      Image(systemName: "square.stack.3d.up")
    }
    .popover(isPresented: $showingPopover){

    Rectangle()

      .frame(width: 500, height: 500)

    }
  }
}

struct Popover_Previews: PreviewProvider {
    static var previews: some View {
        Popover()
        .colorScheme(.dark)
        .previewDevice("iPad Pro (12.9-inch) (3rd generation)")
    }
}

Default behaviour is that is dismisses, once tapped outside.

Question: How can I set the popover to: - Persist (not be dismissed when tapped outside)? - Not block screen when active?

Answer 1

My solution to this problem doesn't involve spinning your own popover lookalike. Simply apply the .interactiveDismissDisabled() modifier to the parent content of the popover, as illustrated in the example below:

import SwiftUI

struct ContentView: View {
    @State private var presentingPopover = false
    @State private var count = 0
    
    var body: some View {
        VStack {
            Button {
                presentingPopover.toggle()
            } label: {
                Text("This view pops!")
            }.popover(isPresented: $presentingPopover) {
                Text("Surprise!")
                    .padding()
                    .interactiveDismissDisabled()
            }.buttonStyle(.borderedProminent)

            Text("Count: \(count)")
            Button {
                count += 1
            } label: {
                Text("Doesn't block other buttons too!")
            }.buttonStyle(.borderedProminent)
        }
        .padding()
    }
}

Tested on iPadOS 16 (Xcode 14.1), demo video included below:

Note: Although it looks like the buttons have lost focus, they are still interact-able, and might be a bug as such behaviour doesn't exist when running on macOS.