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

下面两个函数可以帮助在视图控制器堆栈中找到topViewController。以后可能需要自定义,但是对于这段代码来说,理解topViewController或viewcontroller堆栈的概念非常棒。

- (UIViewController*)findTopViewController {

  id  topControler  = [self topMostController];

  UIViewController* topViewController;
  if([topControler isKindOfClass:[UINavigationController class]]) {
        topViewController = [[(UINavigationController*)topControler viewControllers] lastObject];
   } else if ([topControler isKindOfClass:[UITabBarController class]]) {
        //Here you can get reference of top viewcontroller from stack of viewcontrollers on UITabBarController
  } else {
        //topController is a preented viewController
        topViewController = (UIViewController*)topControler;
  }
    //NSLog(@"Top ViewController is: %@",NSStringFromClass([topController class]));
    return topViewController;
}

- (UIViewController*)topMostController
{
    UIViewController *topController = [UIApplication sharedApplication].keyWindow.rootViewController;

    while (topController.presentedViewController) {
        topController = topController.presentedViewController;
    }
    //NSLog(@"Top View is: %@",NSStringFromClass([topController class]));
    return topController;
}

你可以使用[viewController Class]方法找出一个viewController的类的类型。

其他回答

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!
}

简单的扩展UIApplication在Swift:

注意:

它关心UITabBarController中的moreNavigationController

extension UIApplication {

    class func topViewController(baseViewController: UIViewController? = UIApplication.sharedApplication().keyWindow?.rootViewController) -> UIViewController? {

        if let navigationController = baseViewController as? UINavigationController {
            return topViewController(navigationController.visibleViewController)
        }

        if let tabBarViewController = baseViewController as? UITabBarController {

            let moreNavigationController = tabBarViewController.moreNavigationController

            if let topViewController = moreNavigationController.topViewController where topViewController.view.window != nil {
                return topViewController(topViewController)
            } else if let selectedViewController = tabBarViewController.selectedViewController {
                return topViewController(selectedViewController)
            }
        }

        if let splitViewController = baseViewController as? UISplitViewController where splitViewController.viewControllers.count == 1 {
            return topViewController(splitViewController.viewControllers[0])
        }

        if let presentedViewController = baseViewController?.presentedViewController {
            return topViewController(presentedViewController)
        }

        return baseViewController
    }
}

简单的用法:

if let topViewController = UIApplication.topViewController() {
    //do sth with top view controller
}

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!

不确定这是否会帮助你通过找到最顶层的视图控制器来实现,但我试图呈现一个新的视图控制器,但如果我的根视图控制器已经有一个模态对话框,它会被阻塞,所以我将循环到所有模态视图控制器的顶部使用以下代码:

UIViewController* parentController =[UIApplication sharedApplication].keyWindow.rootViewController;

while( parentController.presentedViewController &&
       parentController != parentController.presentedViewController )
{
    parentController = parentController.presentedViewController;
}

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
}