如何在 Swift 中弹出特定的视图控制器

Question

我使用了下面的 Objective-C 代码来弹出一个特定的 ViewController。

for (UIViewController *controller in self.navigationController.viewControllers) {
    if ([controller isKindOfClass:[AnOldViewController class]]) { 
        //Do not forget to import AnOldViewController.h
        [self.navigationController popToViewController:controller
                                              animated:YES];
        break;
    }
}

我如何在 Swift 中做到这一点？

Answer 1

试试下面的代码：

for controller in self.navigationController!.viewControllers as Array {
    if controller.isKind(of: ViewController.self) {
        self.navigationController!.popToViewController(controller, animated: true)
        break
    }
}

Answer 2

为Swift3+

let viewControllers: [UIViewController] = self.navigationController!.viewControllers
for aViewController in viewControllers {
    if aViewController is YourViewController {
        self.navigationController!.popToViewController(aViewController, animated: true)
    }
}

Answer 3

我更喜欢通用的方式。

我有 UINavigationController 的这个扩展：

extension UINavigationController {

   func backToViewController(vc: Any) {
      // iterate to find the type of vc
      for element in viewControllers as Array {
         if "\(element.dynamicType).Type" == "\(vc.dynamicType)" {
            self.popToViewController(element, animated: true)
            break
         }
      }
   }

}

假设我在导航堆栈中实例化了一个 FOHomeVC class（它是一个 UIViewController）。

所以我会在我的代码中这样做：

self.navigationController?.backToViewController(FOHomeVC.self)

Answer 4

请为 Swift 3.0 使用以下代码：

 let viewControllers: [UIViewController] = self.navigationController!.viewControllers as [UIViewController];

for aViewController:UIViewController in viewControllers {
            if aViewController.isKind(of: YourViewController.self) {
                _ = self.navigationController?.popToViewController(aViewController, animated: true)
            }
        }

Answer 5

Swift 4.0

for controller in self.navigationController!.viewControllers as Array {
        if controller.isKind(of: DashboardVC.self) {
            _ =  self.navigationController!.popToViewController(controller, animated: true)
            break
        }
    }

这很完美。

Answer 6

我在 UINavigationController 中添加了一个扩展，它可以帮助您找到该控制器是否存在于导航堆栈中。如果是，那么它将被弹出到该控制器，否则您将传递新的控制器以使用 pushController 参数进行推送。

extension UINavigationController {

    func containsViewController(ofKind kind: AnyClass) -> Bool {
        return self.viewControllers.contains(where: { [=10=].isKind(of: kind) })
    }

    func popPushToVC(ofKind kind: AnyClass, pushController: UIViewController) {
        if containsViewController(ofKind: kind) {
            for controller in self.viewControllers {
                if controller.isKind(of: kind) {
                    popToViewController(controller, animated: true)
                    break
                }
            }
        } else {
            pushViewController(pushController, animated: true)
        }
    }
}

Answer 7

最近 swift

   @IBAction func popToConversationsVC(_ sender: UIButton) {
        if (self.navigationController != nil) {
            for vc in  self.navigationController!.viewControllers {
                if vc is ConversationsVC {
                     self.navigationController?.popToViewController(vc, animated: false)
                }
            }
            }
    }

Answer 8

从导航堆栈中找到您的视图控制器，如果存在则弹出到该视图控制器

for vc in self.navigationController!.viewControllers {
    if let myViewCont = vc as? VCName 
    {
        self.navigationController?.popToViewController(myViewCont, animated: true)
    }
}

Answer 9

Swift 4 / Swift 5

 for controller in self.navigationController!.viewControllers as Array {
            if controller.isKind(of: HomeViewController.self) {
                self.navigationController!.popToViewController(controller, animated: true)
                break
            }
        }

Answer 10

我需要使用它，因为在某些情况下应用会崩溃：

if let navVC = self.navigationController {
                    let views = navVC.viewControllers as Array
                    for controller in views {
                        if controller.isKind(of: YourVC.self) {
                            navVC.popToViewController(controller, animated: true)
                            return
                        }
                    }
                }

Answer 11

Swift 5

弹出特定 class 的最新实例，例如 SomeViewController:

navigationController?.popToViewController(ofClass: SomeViewController.self)

但您需要添加 UINavigationController 扩展名：

extension UINavigationController {
  func popToViewController(ofClass: AnyClass, animated: Bool = true) {
    if let vc = viewControllers.last(where: { [=11=].isKind(of: ofClass) }) {
      popToViewController(vc, animated: animated)
    }
  }
}

Answer 12

extension UINavigationController {
    func popBack(to vc: AnyClass, animated: Bool = true) {
        guard let elementFound = (viewControllers.filter { [=10=].isKind(of: vc) }).first else {
            fatalError("cannot pop back to \(vc) as it is not in the view hierarchy")
        }
        self.popToViewController(elementFound, animated: animated)
    }
}

Answer 13

没有强制展开的简单且最佳的解决方案是

if let vc = navigationController.viewControllers.filter({[=10=] is YourViewController}).first as? YourViewController {
            self.navigationController.popToViewController(vc, animated: true)
            }

Answer 14

swift5

let controllers : Array = self.navigationController!.viewControllers
self.navigationController!.popToViewController(controllers[1], animated: true)

Answer 15

我更喜欢 "real generic" 和更实用的方法。

所以我想出了以下 UINavigationController 扩展函数。您还可以将第一个函数用于其他用途，您只需要访问导航堆栈中的特定 VC。

扩展

extension UINavigationController {
    func getViewController<T: UIViewController>(of type: T.Type) -> UIViewController? {
        return self.viewControllers.first(where: { [=10=] is T })
    }

    func popToViewController<T: UIViewController>(of type: T.Type, animated: Bool) {
        guard let viewController = self.getViewController(of: type) else { return }
        self.popToViewController(viewController, animated: animated)
    }
}

用法

self.navigationController?.popToViewController(of: YourViewController.self, animated: true)

这应该至少在 Swift 4 和 5 中有效。

Answer 16

Swift 5 @PabloR 的答案在这里：

extension UINavigationController {

   func backToViewController(vc: Any) {
      // iterate to find the type of vc
      for element in viewControllers as Array {
        if "\(type(of: element)).Type" == "\(type(of: vc))" {
            self.popToViewController(element, animated: true)
            break
         }
      }
   }

}

用法：

self.navigationController?.backToViewController(vc: TaskListViewController.self)

Answer 17

对于 Swift 4.0 及更高版本使用过滤器

guard let VC = self.navigationController?.viewControllers.filter({[=10=].isKind(of: YourViewController.self)}).first else {return}
self.navigationController?.popToViewController(VC, animated: true)

Answer 18

这个解决方案对我有用:)

extension UINavigationController {
  func backToViewController(_ viewController: AnyClass, animated: Bool) {

    guard let viewController = self.viewControllers.first(where: {[=10=].isKind(of: viewController)}) else { return }
    self.popToViewController(viewController, animated: animated)
  }
}

Answer 19

我改编自以上所有答案。它看起来像 Yakup Ad 答案，因为它的路很短。

我通过对参数使用泛型来强制类型，确保您必须只将 UIViewController 传递给此函数。
我使用 .first 搜索已经在堆栈中的 viewController。这让我只有一个 VC 然后停止循环。
如果你需要定制一些东西，我也会return传递VC。

让我们尽情享受吧。

extension UINavigationController {

    func popToViewController<T: UIViewController>(_ viewController: T.Type, animated: Bool) -> T? {
        guard let viewController = self.viewControllers.first(where: {[=10=] is T}) else { return nil }
        self.popToViewController(viewController, animated: animated)
        
        return viewController as? T
    }
    
}

用法

let poppedVC = self.navigationController?.popToViewController(HomeViewController.self, animated: true)

如何在 Swift 中弹出特定的视图控制器

How can I pop specific View Controller in Swift

uiviewcontroller

ios

swift