swift4.1非ViewController类实现页面跳转

有时候我们需要直接获取到当前显示的UINavigationController，以便于后续操作，
获取方法如下。此方法可以获取present出来的UINavigationController。

var navigationController : UINavigationController? {
        get {
            var parent: UIViewController?
            if let window = UIApplication.shared.delegate?.window,let rootVC = window?.rootViewController {
                parent = rootVC
                while (parent?.presentedViewController != nil) {
                    parent = parent?.presentedViewController!
                }
                
                if let tabbar = parent as? UITabBarController ,let nav = tabbar.selectedViewController as? UINavigationController {
                    return nav
                }else if let nav = parent as? UINavigationController {
                    return nav
                }
            }
            return nil
        }
}

有时候我们需要在UIView中实现跳转，每次都用block回调很麻烦，那么可以实现一个UIView的Extension

extension UIView {
    open func show(_ vc: UIViewController, sender: Any?) {
        viewController?.show(vc, sender: sender)
    }

    open func present(_ viewControllerToPresent: UIViewController, animated flag: Bool, completion: (() -> Void)? = nil) {
        viewController?.present(viewControllerToPresent, animated: flag, completion: completion)
    }

    public var viewController: UIViewController? {
        var next: UIResponder?
        next = self.next
        repeat {
            if let vc = next as? UIViewController {
                return vc
            } else {
                next = next?.next
            }
        } while next != nil
        return nil
    }
}

有时候我们通过present跳转了多个页面，想要dismiss到根视图：

    private func dismissToRootViewController() {
        var vc: UIViewController = self
        while vc.presentingViewController != nil {
            vc = vc.presentingViewController!
        }
        vc.dismiss(animated: true, completion: nil)
    }