使用选择器通过参数传递的快速调用函数

Question

我想编写一个函数来创建 View ，将它们添加到 stackView 并在单击其中一个项目时调用回调函数:这是一个简化的示例:

func createSelection(callBack:(Int)->()) {
    for i in 0...5 {
        let view = UIView()
        // configure and fill view...
        stackView.addArrangedSubview(view)
        // add gesture recognizer
    }
}

我的问题是，当其中一个 View 被单击时，我不知道如何调用回调函数。我不能在它上面使用选择器，因为它没有声明为 objc 函数。我该怎么做？

Answer 1

您可以为 UIView 创建一个协议(protocol)，并将其用于您喜欢的任何 View 。

import UIKit

protocol Tappable {
    var callback: ((Int) -> ())? { get set }
    var callbackValue: Int? { get set }
    func setupTapGestureRecognizer()
}

class TappableView: UIView, Tappable {
    var callback: ((Int) -> ())?
    var callbackValue: Int?

    func setupTapGestureRecognizer() {
        let tapGestureRecognizer = UITapGestureRecognizer(target: self, action: #selector(tapGestureAction))
        self.addGestureRecognizer(tapGestureRecognizer)
    }

    @objc func tapGestureAction() {
        print("tapped")
        guard let callback = callback, let callbackValue = callbackValue else { return }

        callback(callbackValue)
    }
}

func createSelection(callBack: @escaping (Int) -> ()) {
    for i in 0...5 {
        let view = TappableView()
        view.callback = callBack
        view.callbackValue = i
        view.setupTapGestureRecognizer()
        // Mocking tap action, remove this when you're done:
        view.tapGestureAction()
    }
}

createSelection { (value) in
    print("hi from callback, with value: \(value)")
}

Playground 中的结果:

tapped
hi from callback, with value: 0
tapped
hi from callback, with value: 1
tapped
hi from callback, with value: 2
tapped
hi from callback, with value: 3
tapped
hi from callback, with value: 4
tapped
hi from callback, with value: 5