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

简单的扩展UIApplication在Swift:

注意:

它关心UITabBarController中的moreNavigationController

extension UIApplication {

    class func topViewController(baseViewController: UIViewController? = UIApplication.sharedApplication().keyWindow?.rootViewController) -> UIViewController? {

        if let navigationController = baseViewController as? UINavigationController {
            return topViewController(navigationController.visibleViewController)
        }

        if let tabBarViewController = baseViewController as? UITabBarController {

            let moreNavigationController = tabBarViewController.moreNavigationController

            if let topViewController = moreNavigationController.topViewController where topViewController.view.window != nil {
                return topViewController(topViewController)
            } else if let selectedViewController = tabBarViewController.selectedViewController {
                return topViewController(selectedViewController)
            }
        }

        if let splitViewController = baseViewController as? UISplitViewController where splitViewController.viewControllers.count == 1 {
            return topViewController(splitViewController.viewControllers[0])
        }

        if let presentedViewController = baseViewController?.presentedViewController {
            return topViewController(presentedViewController)
        }

        return baseViewController
    }
}

简单的用法:

if let topViewController = UIApplication.topViewController() {
    //do sth with top view controller
}

其他回答

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
}

另一个解决方案依赖于responder链,它可能工作,也可能不工作,这取决于第一个responder是什么:

获取第一个响应器。 获取与第一个responder相关联的UIViewController。

示例伪代码:

+ (UIViewController *)currentViewController {
    UIView *firstResponder = [self firstResponder]; // from the first link above, but not guaranteed to return a UIView, so this should be handled more appropriately.
    UIViewController *viewController = [firstResponder viewController]; // from the second link above
    return viewController;
}

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]; 

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];

不确定这是否会帮助你通过找到最顶层的视图控制器来实现,但我试图呈现一个新的视图控制器,但如果我的根视图控制器已经有一个模态对话框,它会被阻塞,所以我将循环到所有模态视图控制器的顶部使用以下代码:

UIViewController* parentController =[UIApplication sharedApplication].keyWindow.rootViewController;

while( parentController.presentedViewController &&
       parentController != parentController.presentedViewController )
{
    parentController = parentController.presentedViewController;
}