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

最新Swift版本: 创建一个文件,命名为UIWindowExtension.swift,并粘贴以下代码片段:

import UIKit

public extension UIWindow {
    public var visibleViewController: UIViewController? {
        return UIWindow.getVisibleViewControllerFrom(self.rootViewController)
    }

    public static func getVisibleViewControllerFrom(_ vc: UIViewController?) -> UIViewController? {
        if let nc = vc as? UINavigationController {
            return UIWindow.getVisibleViewControllerFrom(nc.visibleViewController)
        } else if let tc = vc as? UITabBarController {
            return UIWindow.getVisibleViewControllerFrom(tc.selectedViewController)
        } else {
            if let pvc = vc?.presentedViewController {
                return UIWindow.getVisibleViewControllerFrom(pvc)
            } else {
                return vc
            }
        }
    }
}

func getTopViewController() -> UIViewController? {
    let appDelegate = UIApplication.shared.delegate
    if let window = appDelegate!.window {
        return window?.visibleViewController
    }
    return nil
}

可以在任何地方使用:

if let topVC = getTopViewController() {

}

其他回答

一个完整的非递归版本,照顾到不同的场景:

视图控制器正在呈现另一个视图 视图控制器是一个UINavigationController 视图控制器是一个UITabBarController

objective - c

 UIViewController *topViewController = self.window.rootViewController;
 while (true)
 {
     if (topViewController.presentedViewController) {
         topViewController = topViewController.presentedViewController;
     } else if ([topViewController isKindOfClass:[UINavigationController class]]) {
         UINavigationController *nav = (UINavigationController *)topViewController;
         topViewController = nav.topViewController;
     } else if ([topViewController isKindOfClass:[UITabBarController class]]) {
         UITabBarController *tab = (UITabBarController *)topViewController;
         topViewController = tab.selectedViewController;
     } else {
         break;
     }
 }

斯威夫特 4+

extension UIWindow {
    func topViewController() -> UIViewController? {
        var top = self.rootViewController
        while true {
            if let presented = top?.presentedViewController {
                top = presented
            } else if let nav = top as? UINavigationController {
                top = nav.visibleViewController
            } else if let tab = top as? UITabBarController {
                top = tab.selectedViewController
            } else {
                break
            }
        }
        return top
    }
}

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
}

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。

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;
}

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()