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

这对于从任何根视图控件中找到top viewcontroller1非常有效

+ (UIViewController *)topViewControllerFor:(UIViewController *)viewController
{
    if(!viewController.presentedViewController)
        return viewController;
    return [MF5AppDelegate topViewControllerFor:viewController.presentedViewController];
}

/* View Controller for Visible View */

AppDelegate *app = [UIApplication sharedApplication].delegate;
UIViewController *visibleViewController = [AppDelegate topViewControllerFor:app.window.rootViewController]; 

其他回答

这是对Eric的回答的改进:

UIViewController *_topMostController(UIViewController *cont) {
    UIViewController *topController = cont;

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

    if ([topController isKindOfClass:[UINavigationController class]]) {
        UIViewController *visible = ((UINavigationController *)topController).visibleViewController;
        if (visible) {
            topController = visible;
        }
    }

    return (topController != cont ? topController : nil);
}

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

    UIViewController *next = nil;

    while ((next = _topMostController(topController)) != nil) {
        topController = next;
    }

    return topController;
}

UIViewController *cont是一个辅助函数。

现在你所需要做的就是调用topMostController()和最顶端的UIViewController应该被返回!

你可以通过使用找到最顶层视图控制器

NSArray *arrViewControllers=[[self navigationController] viewControllers];
UIViewController *topMostViewController=(UIViewController *)[arrViewControllers objectAtIndex:[arrViewControllers count]-1];

为了避免很多复杂性,我通过在委托中创建一个viewController来跟踪当前的viewController,并在每个viewDidLoad方法中设置它为self,这样每当你加载一个新视图时,委托中持有的viewController将对应于该视图的viewController。这可能是丑陋的,但它工作得很好,没有必要有一个导航控制器或任何废话。

Swift 4.2扩展


extension UIApplication {

    class func topViewController(controller: UIViewController? = UIApplication.shared.keyWindow?.rootViewController) -> UIViewController? {
        if let navigationController = controller as? UINavigationController {
            return topViewController(controller: navigationController.visibleViewController)
        }
        if let tabController = controller as? UITabBarController {
            if let selected = tabController.selectedViewController {
                return topViewController(controller: selected)
            }
        }
        if let presented = controller?.presentedViewController {


            return topViewController(controller: presented)
        }
        return controller
    }
}

在任何地方都可以使用,

 UIApplication.topViewController()?.present(yourController, animated: true, completion: nil)

或者像,

 UIApplication.topViewController()?
                    .navigationController?
                    .popToViewController(yourController,
                                         animated: true)

适合任何类,如UINavigationController, UITabBarController

享受吧!

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
    }