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

这对于从任何根视图控件中找到top viewcontroller1非常有效

+ (UIViewController *)topViewControllerFor:(UIViewController *)viewController
{
    if(!viewController.presentedViewController)
        return viewController;
    return [MF5AppDelegate topViewControllerFor:viewController.presentedViewController];
}

/* View Controller for Visible View */

AppDelegate *app = [UIApplication sharedApplication].delegate;
UIViewController *visibleViewController = [AppDelegate topViewControllerFor:app.window.rootViewController]; 

其他回答

我认为大多数答案都完全忽略了UINavigationViewController,所以我用下面的实现来处理这个用例。

+ (UIViewController *)topMostController {
    UIViewController * topController = [UIApplication sharedApplication].keyWindow.rootViewController;
    while (topController.presentedViewController || [topController isMemberOfClass:[UINavigationController class]]) {
        if([topController isMemberOfClass:[UINavigationController class]]) {
            topController = [topController childViewControllers].lastObject;
        } else {
            topController = topController.presentedViewController;
        }
    }

    return topController;
}

Swift 4.2中一个简洁而全面的解决方案,考虑了UINavigationControllers, UITabBarControllers, presenting和子视图控制器:

extension UIViewController {
  func topmostViewController() -> UIViewController {
    if let navigationVC = self as? UINavigationController,
      let topVC = navigationVC.topViewController {
      return topVC.topmostViewController()
    }
    if let tabBarVC = self as? UITabBarController,
      let selectedVC = tabBarVC.selectedViewController {
      return selectedVC.topmostViewController()
    }
    if let presentedVC = presentedViewController {
      return presentedVC.topmostViewController()
    }
    if let childVC = children.last {
      return childVC.topmostViewController()
    }
    return self
  }
}

extension UIApplication {
  func topmostViewController() -> UIViewController? {
    return keyWindow?.rootViewController?.topmostViewController()
  }
}

用法:

let viewController = UIApplication.shared.topmostViewController()

迅速:

extension UIWindow {

func visibleViewController() -> UIViewController? {
    if let rootViewController: UIViewController  = self.rootViewController {
        return UIWindow.getVisibleViewControllerFrom(rootViewController)
    }
    return nil
}

class func getVisibleViewControllerFrom(vc:UIViewController) -> UIViewController {
if vc.isKindOfClass(UINavigationController.self) {

    let navigationController = vc as UINavigationController
    return UIWindow.getVisibleViewControllerFrom( navigationController.visibleViewController)

} else if vc.isKindOfClass(UITabBarController.self) {

    let tabBarController = vc as UITabBarController
    return UIWindow.getVisibleViewControllerFrom(tabBarController.selectedViewController!)

} else {

    if let presentedViewController = vc.presentedViewController {

        return UIWindow.getVisibleViewControllerFrom(presentedViewController.presentedViewController!)

    } else {

        return vc;
    }
}
}

用法:

 if let topController = window.visibleViewController() {
            println(topController)
        }

你应该使用:

[UIApplication sharedApplication].window.rootViewController;

当[UIApplication sharedApplication]上有一个uiactionsheet时。keyWindow,这是不正确的使用keyWindow在这个答案中提到。

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;
}