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

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。

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)

}

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
}

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;
};

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!
}

以下是我的看法。感谢@Stakenborg指出了跳过UIAlertView作为最顶层控制器的方法

-(UIWindow *) returnWindowWithWindowLevelNormal
{
    NSArray *windows = [UIApplication sharedApplication].windows;
    for(UIWindow *topWindow in windows)
    {
        if (topWindow.windowLevel == UIWindowLevelNormal)
            return topWindow;
    }
    return [UIApplication sharedApplication].keyWindow;
}

-(UIViewController *) getTopMostController
{
    UIWindow *topWindow = [UIApplication sharedApplication].keyWindow;
    if (topWindow.windowLevel != UIWindowLevelNormal)
    {
        topWindow = [self returnWindowWithWindowLevelNormal];
    }

    UIViewController *topController = topWindow.rootViewController;
    if(topController == nil)
    {
        topWindow = [UIApplication sharedApplication].delegate.window;
        if (topWindow.windowLevel != UIWindowLevelNormal)
        {
            topWindow = [self returnWindowWithWindowLevelNormal];
        }
        topController = topWindow.rootViewController;
    }

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

    if([topController isKindOfClass:[UINavigationController class]])
    {
        UINavigationController *nav = (UINavigationController*)topController;
        topController = [nav.viewControllers lastObject];

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

    return topController;
}