ios - 如何在运行时更改 UIMenu 内 UIAction 的状态？

Question

如何更改 UIAction 的状态？目标是在 UIMenu 内的 UIAction 旁边切换状态复选标记。

更改 UIAction 的 state通过存储在 View Controller 中的引用似乎根本没有改变状态。我错过了什么吗？

// View Controller
internal var menuAction: UIAction!

private func generatePullDownMenu() -> UIMenu {
    menuAction = UIAction(
        title: "Foo",
        image: UIImage(systemName: "chevron.down"),
        identifier: UIAction.Identifier("come.sample.action"),
        state:  .on
    ) { _ in self.menuAction.state = .off } // <--- THIS LINE


    let menu = UIMenu(
        title: "Sample Menu",
        image: nil,
        identifier: UIMenu.Identifier("com.sample.menu"),
        options: [],
        children: [menuAction]
    )

    return menu
}

// Inside UI setup code block
let buttonItem = UIBarButtonItem(
    title: "",
    image: UIImage(systemName: "chevron.down"),
    primaryAction: nil,
    menu: generatePullDownMenu()
)

试图改变 action直接从闭包状态并得到“操作是不可变的，因为它是菜单的子项”错误。现在我怀疑一个 Action 对象总是一个不可变的对象。

menuAction = UIAction(
    title: "Foo",
    image: UIImage(systemName: "chevron.down"),
    identifier: UIAction.Identifier("come.sample.action"),
    state:  .on
) { action in action.state = .off } // <--- THIS LINE

Answer 1

更换整个UIMenu状态更改的对象可以解决问题。

// view controller
internal var barButton: UIBarButtonItem!

// UI setup function
barButton = UIBarButtonItem(
    image: UIImage(systemName: "arrow.up.arrow.down.square"),
    primaryAction: nil,
    menu: generatePullDownMenu()
)

// On state change inside UIAction 
let actionNextSeen = UIAction(
    title: "foo",
    image: UIImage(systemName: "hourglass", )
    state: someVariable ? .off : .on
) { _ in
    someVariable = false
    self.barButton.menu = self.generatePullDownMenu()
}

引用
https://developer.apple.com/forums/thread/653862