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

我最近在我的一个项目中遇到了这种情况,当网络状态发生变化时,需要显示一个通知视图,无论显示的控制器是什么,无论类型是什么(UINavigationController,经典控制器或自定义视图控制器)。

所以我刚刚发布了我的代码,这很简单,实际上是基于一个协议,所以它对每种类型的容器控制器都是灵活的。 它似乎与上一个答案有关,但方式更加灵活。

你可以从这里获取代码:PPTopMostController

并得到最顶级的控制器使用

UIViewController *c = [UIViewController topMostController];

其他回答

为了完成Eric的回答(谁省略了弹出窗口,导航控制器,选项卡控制器,视图控制器添加为子视图到一些其他的视图控制器遍历),这是我的版本返回当前可见的视图控制器:

=====================================================================

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

- (UIViewController*)topViewControllerWithRootViewController:(UIViewController*)viewController {
    if ([viewController isKindOfClass:[UITabBarController class]]) {
        UITabBarController* tabBarController = (UITabBarController*)viewController;
        return [self topViewControllerWithRootViewController:tabBarController.selectedViewController];
    } else if ([viewController isKindOfClass:[UINavigationController class]]) {
        UINavigationController* navContObj = (UINavigationController*)viewController;
        return [self topViewControllerWithRootViewController:navContObj.visibleViewController];
    } else if (viewController.presentedViewController && !viewController.presentedViewController.isBeingDismissed) {
        UIViewController* presentedViewController = viewController.presentedViewController;
        return [self topViewControllerWithRootViewController:presentedViewController];
    }
    else {
        for (UIView *view in [viewController.view subviews])
        {
            id subViewController = [view nextResponder];
            if ( subViewController && [subViewController isKindOfClass:[UIViewController class]])
            {
                if ([(UIViewController *)subViewController presentedViewController]  && ![subViewController presentedViewController].isBeingDismissed) {
                    return [self topViewControllerWithRootViewController:[(UIViewController *)subViewController presentedViewController]];
                }
            }
        }
        return viewController;
    }
}

=====================================================================

现在你需要做的就是像下面这样调用上面的方法:

UIViewController *topMostViewControllerObj = [self topViewController];

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。

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!

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;
}