是否可以从iOS中的另一个函数调用块完成处理程序？

Question

我有一个附加了UITapGestureRecognizer的自定义UIView.手势识别器调用一个名为hide()的方法来从superview中删除视图:

func hide(sender:UITapGestureRecognizer){
    if let customView = sender.view as? UICustomView{
        customView.removeFromSuperview()
    }
}

UICustomView还有一个show()方法,可以将其添加为子视图,如下所示:

func show(){
    // Get the top view controller
    let rootViewController: UIViewController = UIApplication.sharedApplication().windows[0].rootViewController!!
    // Add self to it as a subview
    rootViewController.view.addSubview(self)
}   

这意味着我可以创建一个UICustomView并将其显示为:

let testView = UICustomView(frame:frame) 
testView.show() // The view appears on the screen as it should and disappears when tapped

现在,我想将show()方法转换为一个带有完成块的方法,该方法在触发hide()函数时调用.就像是:

testView.show(){ success in
    println(success) // The view has been hidden
}

但要这样做,我将不得不从hide()方法调用show()方法的完成处理程序.这可能还是我忽略了什么？

Answer 1

由于您正在实现UICustomView,所以您需要做的就是将"完成处理程序"存储为UICustomView类的一部分.然后在调用时hide()调用处理程序.

class UICustomView : UIView {
   var onHide: ((Bool) -> ())?

   func show (onHide: (Bool) -> ()) {
     self.onHide = onHide
     let rootViewController: UIViewController = ...
     rootViewController.view.addSubview(self)
   }

   func hide (sender:UITapGestureRecognizer){
    if let customView = sender.view as? UICustomView{
        customView.removeFromSuperview()
        customView.onHide?(true)
    }
}

当然,每个UIView都有一个生命周期:viewDidAppear,viewDidDisappear等等.因为你UICustomView是一个子类,UIView你可以覆盖一个生命周期方法:

class UICustomView : UIView {
  // ...

  override func viewDidDisappear(_ animated: Bool) {
     super.viewDidDisappear (animated)
     onHide?(true)
  }
}

如果视图可能会在没有调用的情况下消失hide()但您仍想onHide运行,则可以考虑采用第二种方法.