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

我的问题有点不同,我在我的应用程序中使用SWRevealViewController。 我使用了Yuchen Zhong的答案,但它总是返回topViewController作为SWRevealViewController。对于那些使用SWRevealViewController或其他一些pod来开发侧菜单的人。以下是我对钟宇晨回答的延伸:

extension UIApplication {
class func topViewController() -> UIViewController? {
    var topVC = shared.keyWindow!.rootViewController
    while true {
        if let presented = topVC?.presentedViewController {
            topVC = presented
        } else if let nav = topVC as? UINavigationController {
            topVC = nav.visibleViewController
        } else if let tab = topVC as? UITabBarController {
            topVC = tab.selectedViewController
        }else if let swRVC = topVC as? SWRevealViewController {
            topVC = swRVC.frontViewController
        } else {
            break
        }
    }
    return topVC
}
}

其他回答

很多答案都是不完整的。虽然这是在Objective-C中,但这是我现在能把它们放在一起的最好的编译,作为一个非递归的块:

链接到Gist,以防修改:https://gist.github.com/benguild/0d149bb3caaabea2dac3d2dca58c0816 供参考/比较的代码:

UIViewController *(^topmostViewControllerForFrontmostNormalLevelWindow)(void) = ^UIViewController *{
    // NOTE: Adapted from various stray answers here:
    //   https://stackoverflow.com/questions/6131205/iphone-how-to-find-topmost-view-controller/20515681

    UIViewController *viewController;

    for (UIWindow *window in UIApplication.sharedApplication.windows.reverseObjectEnumerator.allObjects) {
        if (window.windowLevel == UIWindowLevelNormal) {
            viewController = window.rootViewController;
            break;
        }
    }

    while (viewController != nil) {
        if ([viewController isKindOfClass:[UITabBarController class]]) {
            viewController = ((UITabBarController *)viewController).selectedViewController;
        } else if ([viewController isKindOfClass:[UINavigationController class]]) {
            viewController = ((UINavigationController *)viewController).visibleViewController;
        } else if (viewController.presentedViewController != nil && !viewController.presentedViewController.isBeingDismissed) {
            viewController = viewController.presentedViewController;
        } else if (viewController.childViewControllers.count > 0) {
            viewController = viewController.childViewControllers.lastObject;
        } else {
            BOOL repeat = NO;

            for (UIView *view in viewController.view.subviews.reverseObjectEnumerator.allObjects) {
                if ([view.nextResponder isKindOfClass:[UIViewController class]]) {
                    viewController = (UIViewController *)view.nextResponder;

                    repeat = YES;
                    break;
                }
            }

            if (!repeat) {
                break;
            }
        }
    }

    return viewController;
};

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!

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
}

Swift替代解决方案:

static func topMostController() -> UIViewController {
    var topController = UIApplication.sharedApplication().keyWindow?.rootViewController
    while (topController?.presentedViewController != nil) {
        topController = topController?.presentedViewController
    }

    return topController!
}