当我的代码被分成类时,我在使用自来水手势时遇到了一些麻烦。我之前已将这一切都包含在一个文件中并且运行顺利,因此我假设我在以下代码中做错了一些:
placeContainerView.userInteractionEnabled = true
let showFullPlaceContainerView = UITapGestureRecognizer(target: self, action: Selector(self.showFullPlaceContainerViewFunction(placeContainerView)))
placeContainerView.addGestureRecognizer(showFullPlaceContainerView)
函数showFullContainerViewFunction(placeContainerView)
func showFullPlaceContainerViewFunction(placeContainerView: PlaceContainerView) {
placeContainerView.animateExpandContractContainer()
}
和
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)
}
}
以某种方式,placeContainerView无法识别点击,并且在点击时不会返回任何打印语句。
有什么想法吗?谢谢你的帮助!
答案 0 :(得分:2)
查看placeContainerView及其superview框架和userInteractionEnable。
答案 1 :(得分:2)
按照建议更改了选择器语法
let showFullPlaceContainerView = UITapGestureRecognizer(target: self, action: #selector(self.showFullPlaceContainerViewFunction(_:)))
placeContainerView.addGestureRecognizer(showFullPlaceContainerView)
你的方法将是这样的
func showFullPlaceContainerViewFunction(recognizer: UITapGestureRecognizer) {
let placeContainerView = recognizer.view as! PlaceContainerView
placeContainerView.animateExpandContractContainer()
}
答案 2 :(得分:0)
预期函数的参数在哪里?
placeContainerView.userInteractionEnabled = true
let showFullPlaceContainerView = UITapGestureRecognizer(target: self, action: #selector(YourViewController.yourFunction(_:))
placeContainerView.addGestureRecognizer(showFullPlaceContainerView)
您的预期功能应该是这样的:
func yourFunction(tapGestureRecognizer:UITapGestureRecognizer) {
// Do something
}