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Swift 当我从另一个类调用时发送到实例的无法识别的选择器_Swift_Uitapgesturerecognizer - Fatal编程技术网

Swift 当我从另一个类调用时发送到实例的无法识别的选择器

Swift 当我从另一个类调用时发送到实例的无法识别的选择器,swift,uitapgesturerecognizer,Swift,Uitapgesturerecognizer,当我从另一个类调用识别器时,会出现以下错误: 发送到实例的选择器无法识别, 如果在同一类中没有调用错误 var orderViewCard = OrderVC() let tapGestureRecognizer = UITapGestureRecognizer(target: self, action: #selector(orderViewCard.handleCardTap(recognzier:))) orderViewCard.handleArea.addGestureRecogni

当我从另一个类调用识别器时,会出现以下错误: 发送到实例的选择器无法识别, 如果在同一类中没有调用错误

var orderViewCard = OrderVC()

let tapGestureRecognizer = UITapGestureRecognizer(target: self, action: #selector(orderViewCard.handleCardTap(recognzier:)))
orderViewCard.handleArea.addGestureRecognizer(tapGestureRecognizer)
在OrderVC中:

@objc
    func handleCardTap(recognzier:UITapGestureRecognizer) {
        switch recognzier.state {
        case .ended:
            animateTransitionIfNeeded(state: nextState, duration: 0.9)
        default:
            break
        }
}

您需要做的是相应地选择目标

     var orderViewCard = OrderVC()

    let tapGestureRecognizer = UITapGestureRecognizer(target: orderViewCard, action: #selector(orderViewCard.handleCardTap))
    orderViewCard.handleArea.addGestureRecognizer(tapGestureRecognizer)

如果它应该在另一个类中调用,那么将
self
指定给
target
是没有意义的。输入nil或其他什么?是的,它可以工作,谢谢)