如何在运行时更改UIAction在UIMenu中的状态？

Question

如何更改UIAction的状态？目标是在UIAction内部UIMenu中切换一个状态标记。

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通过存储在视图控制器中的引用更改UIAction的state似乎根本不改变状态。我有遗漏什么吗？

// 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()
)

尝试从闭包中直接更改action状态，并得到"Action，因为它是菜单的子程序“的错误。现在，我怀疑动作对象总是不可变的对象。

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

在状态更改时替换整个UIMenu对象就可以了。

// 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()
}

参考

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

Answer 2

您需要重新创建菜单。此示例还将正确选择被点击的项：

private func setupViews()
    timeFrameButton = UIBarButtonItem(
        image: UIImage(systemName: "calendar"),
        menu: createMenu()
    )
    navigationItem.leftBarButtonItem = timeFrameButton
}

private func createMenu(actionTitle: String? = nil) -> UIMenu {
    let menu = UIMenu(title: "Menu", children: [
        UIAction(title: "Yesterday") { [unowned self] action in
            self.timeFrameButton.menu = createMenu(actionTitle: action.title)
        },
        UIAction(title: "Last week") { [unowned self] action in
            self.timeFrameButton.menu = createMenu(actionTitle: action.title)
        },
        UIAction(title: "Last month") { [unowned self] action in
            self.timeFrameButton.menu = createMenu(actionTitle: action.title)
        }
    ])
    
    if let actionTitle = actionTitle {
        menu.children.forEach { action in
            guard let action = action as? UIAction else {
                return
            }
            if action.title == actionTitle {
                action.state = .on
            }
        }
    } else {
        let action = menu.children.first as? UIAction
        action?.state = .on
    }
    
    return menu
}

Answer 3

不要试图在菜单显示时更改它。通过更改数据来响应选择。同时，菜单会消失，因为用户已经选择了一个操作。但是现在，当用户下次显示菜单时，您可以使用这些数据来构造菜单。