我在Xcode中创建一个Swift Playground，我将TapGestureRecognizer放到UILabel上有问题......
class MyViewController : UIViewController {
override func loadView() {
func tapped() {
print("The label was tapped!")
}
let label11 = UILabel()
label11.frame = CGRect(x: -350, y: 360, width: 350, height: 20)
label11.text = "name."
label11.textColor = .black
label11.isUserInteractionEnabled = true
view.addSubview(label11)
let tap = UITapGestureRecognizer(target: self, action: "tapped")
label11.addGestureRecognizer(tap)
}
}
答案 0 :(得分：1)
您似乎是以编程方式创建视图控制器，因为您只在loadView（正确地没有调用super）中操作，这需要您创建实际{{1}那个控制器。
view或者，您可以使用import UIKit
import PlaygroundSupport
class MyViewController: UIViewController {
override func loadView() {
view = UIView()
let label11 = UILabel()
label11.frame = CGRect(x: 0, y: 360, width: 350, height: 20)
label11.text = "name."
label11.textColor = .yellow
label11.isUserInteractionEnabled = true
view.addSubview(label11)
let tap = UITapGestureRecognizer(target: self, action: #selector(tapped))
label11.addGestureRecognizer(tap)
}
@objc func tapped() {
print("The label was tapped!")
}
}
PlaygroundPage.current.liveView = MyViewController()
复制非程序化视图控制器（这需要您调用super）。
viewDidLoad此外，您的import UIKit
import PlaygroundSupport
class MyViewController: UIViewController {
override func viewDidLoad() {
super.viewDidLoad()
let label11 = UILabel()
label11.frame = CGRect(x: 0, y: 360, width: 350, height: 20)
label11.text = "name."
label11.textColor = .yellow
label11.isUserInteractionEnabled = true
view.addSubview(label11)
let tap = UITapGestureRecognizer(target: self, action: #selector(tapped))
label11.addGestureRecognizer(tap)
}
@objc func tapped() {
print("The label was tapped!")
}
}
PlaygroundPage.current.liveView = MyViewController()
值超出界限，这就是您可能没有看到标签的原因，并且您缺少Swift 4中所需的origin-x语法。