我的自定义UIView未从SuperView中删除

Question

我有一个PopUpView ，我添加到ViewController 。

我创建了一个委托方法didTapOnOKPopUp（），这样当我点击PopUpView中的Ok Button时，它应该从使用它的委托的ViewController中删除。

这是PopUpView.Swift的代码

protocol PopUpViewDelegate: class {
    func didTapOnOKPopUp()
}


class PopUpView: UIView {
weak var delegate : PopUpViewDelegate?

@IBAction func btnOkPopUpTap(_ sender: UIButton)
    {
        delegate?.didTapOnOKPopUp()
    }
}

这是我使用委托方法的ForgotPasswordViewController的代码。

class ForgotPasswordViewController: UIViewController, PopUpViewDelegate {

// I have created an Instance for the PopUpView and assign Delegate also.

func popUpInstance() -> UIView {
        let popUpView = UINib(nibName: "PopUpView", bundle: nil).instantiate(withOwner: nil, options: nil).first as! PopUpView
        popUpView.delegate = self
        return popUpView
    }
// Here I am adding my view as Subview. It's added successfully.
@IBAction func btnSendTap(_ sender: UIButton) {
        self.view.addSubview(self.popUpInstance())
    }

// But when I tapping on OK Button. My PopUpView is not removing from it's View Controller. 

func didTapOnOKPopUp() {

        self.popUpInstance().removeFromSuperview()
    }
}

我试过这个但没有成功！ 请帮我。 谢谢！

Answer 1

每次调用popupinstance()都会创建一个新的PopUp View。

您可以创建对创建的弹出窗口的引用：

private var displayedPopUp: UIView?
@IBAction func btnSendTap(_ sender: UIButton) {
    displayedPopUp = self.popUpInstance()
    self.view.addSubview(displayedPopUp)
}


func didTapOnOKPopUp() {
    self.displayedPopUp?.removeFromSuperview()
    displayedPopUp = nil
}

但我认为在你的情况下使用lazy var更好：

更换

func popUpInstance() -> UIView {
        let popUpView = UINib(nibName: "PopUpView", bundle: nil).instantiate(withOwner: nil, options: nil).first as! PopUpView
        popUpView.delegate = self
        return popUpView
    }

通过：

lazy var popUpInstance : UIView =  {
        let popUpView = UINib(nibName: "PopUpView", bundle: nil).instantiate(withOwner: nil, options: nil).first as! PopUpView
        popUpView.delegate = self
        return popUpView
    }()

现在每次调用popUpInstance都会返回相同的popUp实例

Answer 2

每次调用.popUpInstance() ，它都会创建一个全新的PopupView实例，从而导致您丢失对视图层次结构中先前创建和添加的实例的引用。

将popUpView定义为实例变量，你应该好好去：

class ForgotPasswordViewController: UIViewController, PopUpViewDelegate {

  private lazy var popupView: PopUpView = {
    let popUpView = UINib(nibName: "PopUpView", bundle: nil)
      .instantiate(withOwner: nil, options: nil)
      .first as! PopUpView

     popUpView.delegate = self
     return popUpView
  }()

  @IBAction func btnSendTap(_ sender: UIButton) {
    self.view.addSubview(self.popupView)
  }

  func didTapOnOKPopUp() {
    self.popupView.removeFromSuperview()
  }
}

Answer 3

每次调用popUpInstance函数时，都会创建另一个PopUpView实例，当你这样做时，你的委托就不相关了。

你可以通过某些方式完成这部分代码：

1. 创建popUpInstance（）函数并将实例保存为类参数

2. 像这样制作一个类参数

    private lazy var popupView: PopUpView = { let popUpView = UINib(nibName: "PopUpView", bundle: nil).instantiate(withOwner: nil, options: nil).first as! PopUpView popUpView.delegate = self return popUpView }()