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

这是一个Swift的应用程序的实现与UINavigationController的根。

  if let nav = UIApplication.sharedApplication().keyWindow?.rootViewController as? UINavigationController{
        //get the current's navigation view controller
        var vc = nav.topViewController
        while vc?.presentedViewController != nil {
            vc = vc?.presentedViewController
        }
        return vc
    }

其他回答

为了完成JonasG的回答(谁在遍历时遗漏了标签栏控制器),这里是我返回当前可见的视图控制器的版本:

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

- (UIViewController*)topViewControllerWithRootViewController:(UIViewController*)rootViewController {
    if ([rootViewController isKindOfClass:[UITabBarController class]]) {
        UITabBarController* tabBarController = (UITabBarController*)rootViewController;
        return [self topViewControllerWithRootViewController:tabBarController.selectedViewController];
    } else if ([rootViewController isKindOfClass:[UINavigationController class]]) {
        UINavigationController* navigationController = (UINavigationController*)rootViewController;
        return [self topViewControllerWithRootViewController:navigationController.visibleViewController];
    } else if (rootViewController.presentedViewController) {
        UIViewController* presentedViewController = rootViewController.presentedViewController;
        return [self topViewControllerWithRootViewController:presentedViewController];
    } else {
        return rootViewController;
    }
}

另一个Swift解决方案

extension UIViewController {
    static var topmostViewController: UIViewController? {
        return UIApplication.sharedApplication().keyWindow?.topmostViewController
    }

    var topmostViewController: UIViewController? {
        return presentedViewController?.topmostViewController ?? self
    }
}

extension UINavigationController {
    override var topmostViewController: UIViewController? {
        return visibleViewController?.topmostViewController
    }
}

extension UITabBarController {
    override var topmostViewController: UIViewController? {
        return selectedViewController?.topmostViewController
    }
}

extension UIWindow {
    var topmostViewController: UIViewController? {
        return rootViewController?.topmostViewController
    }
}

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。

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()

I am thinking that perhaps one thing is being overlooked here. Perhaps it is better to pass the parent viewController into the function that is using the viewController. If you are fishing around in the view hierarchy to find the top view controller that it is probably violating separation of the Model layer and UI layer and is a code smell. Just pointing this out, I did the same, then realized it was much simpler just to pass it in to function, by having the model operation return to the UI layer where I have a reference to the view controller.