How to dismiss popover when click source view

[maxwellyue]

Button("Present popover!") {
   // how to make this 'toggle' working ???
    present.toggle()
}
.popover(present: $present) { 
    Text("Hi, I'm a popover.")
}

When popover is presented, If I click this button again, I want it to dismiss this popover. But in fact, it first dismiss then present again. So how to make this 'toggle' button working?

[tazmancoder]

I have found that using the following works for me.

This will dismiss the popover by tapping outside $0.dismissal.mode = .tapOutside

I have also set an auto dismissal like this: You would attach this onAppear to the popover view .onAppear { DispatchQueue.main.asyncAfter(deadline: .now() + 5, execute: { showPopoverHint.toggle() }) }

All this works for me.

[raysarebest]

Rendering a GeometryReader as a background of the Button and making the popover a modifier of a Spacer in the GeometryReader and setting its excludedFrames to contain the GeometryReader's frame in the .global coordinate space did the trick for me

Button("Present popover!") {
    present.toggle()
}
.background(
    GeometryReader { geometry in
        Spacer()
            .popover(present: $present, attributes: { attributes in
                attributes.dismissal.excludedFrames = {
                    return [geometry.frame(in: .global)]
                }
            }, view: {
                Text("Hi, I'm a popover.")
            })
    }
)

[aehlke]

This needs to be a lib function but thank you for the creative solution!

[aheze]

Nice idea and fix, thanks. Will see what I can do

[aehlke]

Unfortunately I can't find a working solution for accomplishing this with List items and popovers activated by selection, because List first removes the selection before adding it again. I'm now doing it by detecting extremely quick close/reopen on the same item.

[whuxujiyuan]

Button("Present popover!") {
    present.toggle()
}
.allowsHitTesting(!present)
.popover(present: $present) { 
    Text("Hi, I'm a popover.")
}

😂