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)
}
}
您似乎是以编程方式创建视图控制器,因为您只在loadView
中操作(您正确地没有调用super
),这需要您创建该控制器的实际view
。
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()
或者,您可以使用viewDidLoad
(需要您调用super)来复制非编程视图控制器。
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()
此外,您的origin-x
值超出界限,这就是为什么您可能没有看到标签,并且您缺少Swift 4中所需的@objc
语法。