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

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;
 }

其他回答

iOS 4在UIWindow上引入了rootViewController属性:

[UIApplication sharedApplication].keyWindow.rootViewController;

你需要在创建视图控制器后自己设置。

我认为Rajesh的解决方案几乎是完美的,但我认为从上到下遍历子视图更好,我改为如下:

+ (UIViewController *)topViewController:(UIViewController *)viewController{

    if (viewController.presentedViewController)
    {

            UIViewController *presentedViewController = viewController.presentedViewController;
            return [self topViewController:presentedViewController];
     } 
     else if ([viewController isKindOfClass:[UITabBarController class]])
     {

            UITabBarController *tabBarController = (UITabBarController *)viewController;
            return [self topViewController:tabBarController.selectedViewController];
    }

         else if ([viewController isKindOfClass:[UINavigationController class]])
    {   

            UINavigationController *navController = (UINavigationController *)viewController;

            return [self topViewController:navController.visibleViewController];
    }

    // Handling UIViewController's added as subviews to some other views.
    else {

        NSInteger subCount = [viewController.view subviews].count - 1;

        for (NSInteger index = subCount; index >=0 ; --index)
        {

            UIView *view = [[viewController.view subviews] objectAtIndex:index];

            id subViewController = [view nextResponder];    // Key property which most of us are unaware of / rarely use.

            if ( subViewController && [subViewController isKindOfClass:[UIViewController class]])
            {
                return [self topViewController:subViewController];
            }
        }
        return viewController;
    }
}

斯威夫特5

试试这个

let topVisibleVC = UIApplication.shared.keyWindow?.rootViewController?.visibleViewController

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.

这是一个Swift的应用程序的实现与UINavigationController的根。

  if let nav = UIApplication.sharedApplication().keyWindow?.rootViewController as? UINavigationController{
        //get the current's navigation view controller
        var vc = nav.topViewController
        while vc?.presentedViewController != nil {
            vc = vc?.presentedViewController
        }
        return vc
    }