ios - 如何在 Swift 3 中制作可点击的 UILabel？

Question

我正在尝试按照以下代码制作可点击的UILabel:

class ViewNotificationsDetails: UIViewController {    
   @IBOutlet weak var back: UILabel!

      override func viewDidLoad() {
          super.viewDidLoad()

          let tap = UITapGestureRecognizer(target: self, action: #selector(ViewNotificationsDetails.tapFunction))
          back.isUserInteractionEnabled = true
          back.addGestureRecognizer(tap)
      }

      @objc func tapFunction(sender:UITapGestureRecognizer) {
          print("tap working")
      } 
}

但是在执行代码时，我收到错误 ->

Thread 1: Fatal error: Unexpectedly found nil while unwrapping an Optional value on the line "back.isUserInteractionEnabled = true".

可能是什么问题？

Answer 1

试试这个代码，对我来说效果很好

class ViewController: UIViewController {
@IBOutlet weak var cliclableLable: UILabel!

override func viewDidLoad() {
    super.viewDidLoad()

    let tap = UITapGestureRecognizer(target: self, action: #selector(ViewController.tapFunction))
    cliclableLable.isUserInteractionEnabled = true
    cliclableLable.addGestureRecognizer(tap)
}

func tapFunction(sender:UITapGestureRecognizer) {
    print("tap working")
}

}

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