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

另一个Swift解决方案

func topController() -> UIViewController? {

    // recursive follow
    func follow(from:UIViewController?) -> UIViewController? {
        if let to = (from as? UITabBarController)?.selectedViewController {
            return follow(to)
        } else if let to = (from as? UINavigationController)?.visibleViewController {
            return follow(to)
        } else if let to = from?.presentedViewController {
            return follow(to)
        }
        return from
    }

    let root = UIApplication.sharedApplication().keyWindow?.rootViewController

    return follow(root)

}

其他回答

另一个Swift解决方案

func topController() -> UIViewController? {

    // recursive follow
    func follow(from:UIViewController?) -> UIViewController? {
        if let to = (from as? UITabBarController)?.selectedViewController {
            return follow(to)
        } else if let to = (from as? UINavigationController)?.visibleViewController {
            return follow(to)
        } else if let to = from?.presentedViewController {
            return follow(to)
        }
        return from
    }

    let root = UIApplication.sharedApplication().keyWindow?.rootViewController

    return follow(root)

}

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]; 

以下是对我有效的方法。

我发现有时候控制器在键窗口上是nil,因为键窗口是一些操作系统的东西,如警报等。

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

     if (topController == nil)
     {
         // The windows in the array are ordered from back to front by window level; thus,
         // the last window in the array is on top of all other app windows.
         for (UIWindow *aWndow in [[UIApplication sharedApplication].windows reverseObjectEnumerator])
         {
             topController = aWndow.rootViewController;
             if (topController)
                 break;
         }
     }

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

     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;
}

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;
}