扩展@Eric的回答，你需要小心，keyWindow实际上是你想要的窗口。例如，如果您试图在点击警报视图中的某些内容后使用此方法，keyWindow实际上将是警报的窗口，这无疑会给您带来问题。这发生在我在野外通过警报处理深度链接时，并导致SIGABRTs没有堆栈跟踪。要调试的婊子。

下面是我现在使用的代码:

- (UIViewController *)getTopMostViewController {
    UIWindow *topWindow = [UIApplication sharedApplication].keyWindow;
    if (topWindow.windowLevel != UIWindowLevelNormal) {
        NSArray *windows = [UIApplication sharedApplication].windows;
        for(topWindow in windows)
        {
            if (topWindow.windowLevel == UIWindowLevelNormal)
                break;
        }
    }

    UIViewController *topViewController = topWindow.rootViewController;

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

    return topViewController;
}

你可以把这个和你喜欢的从这个问题的其他答案中检索顶视图控制器的方法混合在一起。

- (UIViewController*)topViewController {
    return [self topViewControllerWithRootViewController:[UIApplication sharedApplication].keyWindow.rootViewController];
}

- (UIViewController*)topViewControllerWithRootViewController:(UIViewController*)rootViewController {
    if ([rootViewController isKindOfClass:[UITabBarController class]]) {
        UITabBarController* tabBarController = (UITabBarController*)rootViewController;
        return [self topViewControllerWithRootViewController:tabBarController.selectedViewController];
    } else if ([rootViewController isKindOfClass:[UINavigationController class]]) {
        UINavigationController* navigationController = (UINavigationController*)rootViewController;
        return [self topViewControllerWithRootViewController:navigationController.visibleViewController];
    } else if (rootViewController.presentedViewController) {
        UIViewController* presentedViewController = rootViewController.presentedViewController;
        return [self topViewControllerWithRootViewController:presentedViewController];
    } else {
        return rootViewController;
    }
}

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)

}

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()

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。

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.