Swift:点按手势无法识别

Swift: tap Gesture not recognized

当我的代码被分成 类 时,我在使用点击手势时遇到了一些问题。我之前将所有这些都包含在一个文件中并且运行顺利,所以我假设我在以下代码中做错了:

placeContainerView.userInteractionEnabled = true

let showFullPlaceContainerView = UITapGestureRecognizer(target: self, action: Selector(self.showFullPlaceContainerViewFunction(placeContainerView)))
placeContainerView.addGestureRecognizer(showFullPlaceContainerView)

其中函数 showFullContainerViewFunction(placeContainerView)

func showFullPlaceContainerViewFunction(placeContainerView: PlaceContainerView) {
    placeContainerView.animateExpandContractContainer()   
}

func animateExpandContractContainer() {
    print("Tap gesture working")
    if self.displayingPlaceLabel == false {
        print(self.displayingPlaceLabel)
        self.displayingPlaceLabel = true

        UIView.animateWithDuration(0.4, delay: 0.0, options: [], animations: {
            self.center.x += 180
            }, completion: nil)
    } else {
        self.displayingPlaceLabel = false
        UIView.animateWithDuration(0.4, delay: 0.0, options: [], animations: {
            self.center.x -= 180
            }, completion: nil)
    }

}

placeContainerView 无法识别点击,并且在点击时 return 不打印任何语句。

有什么想法吗?感谢您的帮助!

查看 placeContainerView 及其父视图框架和 userInteractionEnable。

按照您的建议更改了选择器语法

let showFullPlaceContainerView = UITapGestureRecognizer(target: self, action: #selector(self.showFullPlaceContainerViewFunction(_:)))
placeContainerView.addGestureRecognizer(showFullPlaceContainerView)

你的方法是这样的

func showFullPlaceContainerViewFunction(recognizer: UITapGestureRecognizer) {
    let placeContainerView = recognizer.view as! PlaceContainerView
    placeContainerView.animateExpandContractContainer()   
}

期望函数的参数在哪里?

placeContainerView.userInteractionEnabled = true
let showFullPlaceContainerView = UITapGestureRecognizer(target: self, action: #selector(YourViewController.yourFunction(_:))
placeContainerView.addGestureRecognizer(showFullPlaceContainerView)

您期望的功能应该是这样的:

func yourFunction(tapGestureRecognizer:UITapGestureRecognizer) {
   // Do something
}