当返回按钮按下，忽略交互式弹出与屏幕边缘手势。

override func viewWillDisappear(_ animated: Bool) {
    super.viewWillDisappear(animated)
    
    if isMovingFromParent, transitionCoordinator?.isInteractive == false {
      // code here
    }
  }

在我的例子中，viewWillDisappear效果最好。但在某些情况下，必须修改之前的视图控制器。所以这是我的解决方案访问之前的视图控制器，它在Swift 4中工作:

override func viewWillDisappear(_ animated: Bool) {
        super.viewWillDisappear(animated)
        if isMovingFromParentViewController {
            if let viewControllers = self.navigationController?.viewControllers {
                if (viewControllers.count >= 1) {
                    let previousViewController = viewControllers[viewControllers.count-1] as! NameOfDestinationViewController
                    // whatever you want to do
                    previousViewController.callOrModifySomething()
                }
            }
        }
    }

试试这个。

self.navigationItem.leftBarButtonItem?.target = "methodname"
func methodname ( ) {            
  //    enter code here
}

这个也试试。

override func viewWillAppear(animated: Bool) {
  //empty your array
}

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) {

}

我通过以下方法做到了这一点:

斯威夫特3

override func didMoveToParentViewController(parent: UIViewController?) {
   super.didMoveToParentViewController(parent)

   if parent == nil {
      println("Back Button pressed.")
      delegate?.goingBack()
   }           
}

斯威夫特4

override func didMove(toParent parent: UIViewController?) {
    super.didMove(toParent: parent)

    if parent == nil {
        debugPrint("Back Button pressed.")
    }
}

不需要自定义后退按钮。

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.")
    }
}