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我的自定义UIView未从SuperView中删除

uiview swift ios

iHarshil l0gg3r · 技术社区 · 7 年前

弹出视图 我把它加到 视图控制器

我已经创建了一个委托方法 didtaponokpoup() 因此,当我在PopUpView中单击Ok按钮时,它应该从使用它的委托的ViewController中删除。

弹出视图。斯威夫特

protocol PopUpViewDelegate: class {
    func didTapOnOKPopUp()
}


class PopUpView: UIView {
weak var delegate : PopUpViewDelegate?

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

这是你的密码 放弃PasswordViewController 在这里我使用委托方法。

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()
    }
}

我试过了 this 但没有成功!请帮帮我。谢谢

3 回复 | 直到 7 年前

CZ54 7 年前

popupinstance() 创建新的 PopUp 看法

您可以创建对已创建弹出窗口的引用:

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


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

lazy var

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

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

现在每个电话 popUpInstance 将返回与弹出窗口相同的实例

Ozgur Vatansever 7 年前

每次你打电话 .popUpInstance() ,它创造了一个全新的 PopupView 实例,从而导致丢失对视图层次结构中先前创建和添加的引用。

定义 popUpView

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()
  }
}

iHarshil l0gg3r 7 年前

每次调用函数popUpInstance时,都会创建另一个PopUpView实例,这样做时,您的委托就不相关了。

创建popUpInstance()函数并将实例另存为类参数

生成这样的类参数

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 
}()