swift - 如何将 View 作为参数传递(快速)

Question

我正在尝试将功能齐全的滑动手势代码移至view model清理view controller但代码使用了很多self和view引用文献，所以我想我需要传递 view或UIView.self作为调用函数时的参数。但无法让它工作。尝试过:

vm.swipeCode(myView: self.view)

func swipeCode(myView: UIView) {...

但是它崩溃了。经过一番研究后，我还尝试了 inout 的变体和&但无济于事。这是完整的刷卡代码(它引用回view controller，但当事情开始工作时我也会移动它们:))

var myVC = RecipesViewController()

func swipeCode(myView: UIView) {
    //SWIPE RIGHT
    let swipingRight = UISwipeGestureRecognizer()
    swipingRight.addTarget(self, action: #selector(myVC.swipeRight))
    swipingRight.direction = .right
    swipingRight.delegate = self as? UIGestureRecognizerDelegate
    swipingRight.cancelsTouchesInView = false

    myView.isUserInteractionEnabled = true
    myView.addGestureRecognizer(swipingRight)

    //// ALLOW SWIPE LEFT ////
    let swipingLeft = UISwipeGestureRecognizer()
    swipingLeft.addTarget(self, action: #selector(myVC.swipeLeft))
    swipingLeft.direction = .left
    swipingLeft.delegate = self as? UIGestureRecognizerDelegate
    swipingLeft.cancelsTouchesInView = false

    myView.isUserInteractionEnabled = true
    myView.addGestureRecognizer(swipingLeft)
}

Answer 1

崩溃可能是因为这两行:

swipingRight.addTarget(self, action: #selector(myVC.swipeRight))
swipingLeft.addTarget(self, action: #selector(myVC.swipeLeft))

您将 myVC.swipeLeft 作为操作传递，但将 self 作为目标，因此手势识别器将尝试查找 swipeLeft 方法在 self 中，它不存在。

您应该始终确保该操作是目标的成员:

swipingRight.addTarget(myVC, action: #selector(myVC.swipeRight))
swipingLeft.addTarget(myVC, action: #selector(myVC.swipeLeft))