Swift - 防止 UIViewController 中的返回事件

我有一个关于取消 UIViewController 中后退按钮触发的后退事件的问题。在 Objective-C 中有以下内容扩大 https://github.com/onegray/UIViewController-BackButtonHandler。我真的不知道如何将其转换为 swift。我到目前为止尝试的是用我自己的函数覆盖 backBarButton 但它不起作用：

    navigation.backBarButtonItem?.action = #selector(MyController.back)
    navigation.backBarButtonItem?.target = self

我搜索了诸如委托函数之类的东西，但找不到后退按钮的任何内容。

当我遇到这个问题时，我重写了这个扩大 https://github.com/onegray/UIViewController-BackButtonHandler到斯威夫特 3

该解决方案使系统后退按钮带有“

public protocol VCWithBackButtonHandler {
     func shouldPopOnBackButton() -> Bool
}

extension UINavigationController: UINavigationBarDelegate  {
    public func navigationBar(_ navigationBar: UINavigationBar, shouldPop item: UINavigationItem) -> Bool {

        if viewControllers.count < (navigationBar.items?.count) ?? 0 {
            return true
        }

        var shouldPop = true
        let vc = self.topViewController

        if let vc = vc as? VCWithBackButtonHandler {
            shouldPop = vc.shouldPopOnBackButton()
        }

        if shouldPop {
            DispatchQueue.main.async {[weak self] in
                _ = self?.popViewController(animated: true)
            }
        } else {
            for subView in navigationBar.subviews {
                if(0 < subView.alpha && subView.alpha < 1) {
                    UIView.animate(withDuration: 0.25, animations: {
                        subView.alpha = 1
                    })
                }
            }
        }

        return false
    }
}

Usage:

class ViewController: UIViewController,VCWithBackButtonHandler{
    public func shouldPopOnBackButton() -> Bool {
        return false
    }
}