我需要执行一个动作(清空一个数组),当UINavigationController的后退按钮被按下,而按钮仍然导致堆栈上的前一个ViewController出现。我如何使用swift来实现这一点?


当前回答

这是我自己解决问题的方法

override func viewWillAppear(_ animated: Bool) {
    super.viewWillAppear(animated)
    self.navigationItem.leftBarButtonItem?.action = #selector(self.back(sender:))
    self.navigationItem.leftBarButtonItem?.target = self
}

@objc func back(sender: UIBarButtonItem) {

}

其他回答

override func willMove(toParent parent: UIViewController?)
{
    super.willMove(toParent: parent)
    if parent == nil
    {
        print("This VC is 'will' be popped. i.e. the back button was pressed.")
    }
}

如果你正在使用navigationController,那么将UINavigationControllerDelegate协议添加到类中,并添加delegate方法,如下所示:

class ViewController:UINavigationControllerDelegate {

    func navigationController(navigationController: UINavigationController, willShowViewController viewController: UIViewController,
animated: Bool) {
        if viewController === self {
            // do here what you want
        }
    }
}

每当导航控制器滑到新屏幕时,就调用此方法。如果返回按钮被按下,新的视图控制器就是ViewController本身。

在离开电流控制器之前,我需要显示警报。所以我是这样做的:

添加扩展UINavigationController与UINavigationBarDelegate 添加选择器到你的控制器导航

它的工作)

extension UINavigationController: UINavigationBarDelegate {
    public func navigationBar(_ navigationBar: UINavigationBar, shouldPop item: UINavigationItem) -> Bool {
        if let items = navigationBar.items, viewControllers.count < items.count {
            return true
        }

        let clientInfoVC = topViewController as? ClientInfoVC
        if clientInfoVC?.responds(to: #selector(clientInfoVC?.navigationShouldPopOnBack)) ?? false {
            clientInfoVC?.navigationShouldPopOnBack(completion: { isAllowPop in
                if isAllowPop {
                    DispatchQueue.main.async {
                        self.popViewController(animated: true)
                    }
                }
            })
        }

        DispatchQueue.main.async {
            self.popViewController(animated: true)
        }

        return false
    }
}

@objc func navigationShouldPopOnBack(completion: @escaping (Bool) -> ()) {
        let ok = UIAlertAction(title: R.string.alert.actionOk(), style: .default) { _ in
            completion(true)
        }
        let cancel = UIAlertAction(title: R.string.alert.actionCancel(), style: .cancel) { _ in
            completion(false)
        }
        let alertController = UIAlertController(title: "", message: R.string.alert.contractMessage(), preferredStyle: .alert)
        alertController.addAction(ok)
        alertController.addAction(cancel)
        present(alertController, animated: true, completion: nil)
    }

你可以在视图控制器中做一些事情

override func navigationShouldPopOnBackButton() -> Bool {
    self.backAction() //Your action you want to perform.
    return true
}

完整回答使用 当导航条上的“后退”按钮被按下时进行检测

我通过调用/重写viewWillDisappear来实现这一点,然后像这样访问navigationController的堆栈:

override func viewWillDisappear(animated: Bool) {
    super.viewWillDisappear(animated)

    let stack = self.navigationController?.viewControllers.count

    if stack >= 2 {
        // for whatever reason, the last item on the stack is the TaskBuilderViewController (not self), so we only use -1 to access it
        if let lastitem = self.navigationController?.viewControllers[stack! - 1] as? theViewControllerYoureTryingToAccess {
            // hand over the data via public property or call a public method of theViewControllerYoureTryingToAccess, like
            lastitem.emptyArray()
            lastitem.value = 5
        }
    }
}