我遇到了一些情况,现在,它将是方便的,能够找到“最顶层”的视图控制器(一个负责当前视图),但还没有找到一种方法。

基本上,挑战是这样的:给定一个在一个类中执行,这个类不是一个视图控制器(或一个视图)[并且没有活动视图的地址],并且没有传递最顶层视图控制器的地址(或者,比如说,导航控制器的地址),是否有可能找到那个视图控制器?(如果是的话,是怎么做到的?)

或者,如果找不到,有没有可能找到最高处的风景?


当前回答

我知道已经很晚了,可能是多余的。但以下是我提出的对我有用的片段:

    static func topViewController() -> UIViewController? {
        return topViewController(vc: UIApplication.shared.keyWindow?.rootViewController)
    }

    private static func topViewController(vc:UIViewController?) -> UIViewController? {
        if let rootVC = vc {
            guard let presentedVC = rootVC.presentedViewController else {
                return rootVC
            }
            if let presentedNavVC = presentedVC as? UINavigationController {
                let lastVC = presentedNavVC.viewControllers.last
                return topViewController(vc: lastVC)
            }
            return topViewController(vc: presentedVC)
        }
        return nil
    }

其他回答

使用扩展为Swift获得最顶部的视图控制器

代码:

extension UIViewController {
    @objc func topMostViewController() -> UIViewController {
        // Handling Modal views
        if let presentedViewController = self.presentedViewController {
            return presentedViewController.topMostViewController()
        }
        // Handling UIViewController's added as subviews to some other views.
        else {
            for view in self.view.subviews
            {
                // Key property which most of us are unaware of / rarely use.
                if let subViewController = view.next {
                    if subViewController is UIViewController {
                        let viewController = subViewController as! UIViewController
                        return viewController.topMostViewController()
                    }
                }
            }
            return self
        }
    }
}

extension UITabBarController {
    override func topMostViewController() -> UIViewController {
        return self.selectedViewController!.topMostViewController()
    }
}

extension UINavigationController {
    override func topMostViewController() -> UIViewController {
        return self.visibleViewController!.topMostViewController()
    }
}

用法:

UIApplication.sharedApplication().keyWindow!.rootViewController!.topMostViewController()

使用下面的扩展抓取当前可见的UIViewController。适用于Swift 4.0及更高版本

Swift 4.0及以上版本:

extension UIApplication {
    
    class func topViewController(_ viewController: UIViewController? = UIApplication.shared.keyWindow?.rootViewController) -> UIViewController? {
        if let nav = viewController as? UINavigationController {
            return topViewController(nav.visibleViewController)
        }
        if let tab = viewController as? UITabBarController {
            if let selected = tab.selectedViewController {
                return topViewController(selected)
            }
        }
        if let presented = viewController?.presentedViewController {
            return topViewController(presented)
        }
        return viewController
    }
}

如何使用?

let objViewcontroller = UIApplication.topViewController()

我认为你需要一个公认的答案和@fishstix的组合

+ (UIViewController*) topMostController
{
    UIViewController *topController = [UIApplication sharedApplication].keyWindow.rootViewController;

    while (topController.presentedViewController) {
        topController = topController.presentedViewController;
    }

    return topController;
}

Swift 3.0 +

func topMostController() -> UIViewController? {
    guard let window = UIApplication.shared.keyWindow, let rootViewController = window.rootViewController else {
        return nil
    }

    var topController = rootViewController

    while let newTopController = topController.presentedViewController {
        topController = newTopController
    }

    return topController
}

I am thinking that perhaps one thing is being overlooked here. Perhaps it is better to pass the parent viewController into the function that is using the viewController. If you are fishing around in the view hierarchy to find the top view controller that it is probably violating separation of the Model layer and UI layer and is a code smell. Just pointing this out, I did the same, then realized it was much simpler just to pass it in to function, by having the model operation return to the UI layer where I have a reference to the view controller.

最新Swift版本: 创建一个文件,命名为UIWindowExtension.swift,并粘贴以下代码片段:

import UIKit

public extension UIWindow {
    public var visibleViewController: UIViewController? {
        return UIWindow.getVisibleViewControllerFrom(self.rootViewController)
    }

    public static func getVisibleViewControllerFrom(_ vc: UIViewController?) -> UIViewController? {
        if let nc = vc as? UINavigationController {
            return UIWindow.getVisibleViewControllerFrom(nc.visibleViewController)
        } else if let tc = vc as? UITabBarController {
            return UIWindow.getVisibleViewControllerFrom(tc.selectedViewController)
        } else {
            if let pvc = vc?.presentedViewController {
                return UIWindow.getVisibleViewControllerFrom(pvc)
            } else {
                return vc
            }
        }
    }
}

func getTopViewController() -> UIViewController? {
    let appDelegate = UIApplication.shared.delegate
    if let window = appDelegate!.window {
        return window?.visibleViewController
    }
    return nil
}

可以在任何地方使用:

if let topVC = getTopViewController() {

}