ios - 滑动时单个标签更新值

Question

由于一些强大的帮助，我现在可以在各个标签上使用滑动功能了!现在我只需要这些标签中的每一个在滑动时单独更新，因此每个标签都有自己单独的计数器并且不会影响其他标签。

这是我的代码:

导入 UIKit

class ViewController: UIViewController {

    var counter = 0

    @IBOutlet weak var label1: UILabel!
    @IBOutlet weak var label2: UILabel!
    @IBOutlet weak var label3: UILabel!

    var counters: [UILabel: Int] = [:]

    override func viewDidLoad() {
        super.viewDidLoad()

        for label: UILabel in [label1, label2, label3] {

            counters[label] = 0

            for direction: UISwipeGestureRecognizerDirection in [.up, .down, .left, .right] {

                let swipeGesture = UISwipeGestureRecognizer(target: self, action: #selector(didSwipe(_:)))
                swipeGesture.direction = direction
                label.addGestureRecognizer(swipeGesture)
                label.isUserInteractionEnabled = true
                label.isMultipleTouchEnabled = true
            }
        }
    }

    @objc func didSwipe(_ gestureRecognizer: UISwipeGestureRecognizer) {



        guard let label = gestureRecognizer.view as? UILabel else { return }

        debugPrint("\(gestureRecognizer.direction)")

        switch gestureRecognizer.direction {

        case .up:
            counters[label] = counters[label]! + 5
            print(counters)

        case .down:
            counters[label] = 0
            print(counters)

        case .left:
             counters[label] = counters[label]! - 1
            print(counters)

        case .right:
             counters[label] = counters[label]! + 1
            print(counters)

        default:
            label.text = "0"
        }
    }

Answer 1

你需要打电话

label.text = "\(counters[label]!)"

在 switch 语句的末尾。这应该有效。