ios - 在#selector Swift 2.2 中传递参数

Question

我有一个方法:

func followUnfollow(followIcon: UIImageView, channelId: String) {
    let followUnfollow = followIcon
    let tapGestureRecognizer = UITapGestureRecognizer(target: self, action: #selector(self.followIconTapped(_:)))
    followUnfollow.userInteractionEnabled = true
    followUnfollow.addGestureRecognizer(tapGestureRecognizer)
}

我还有一个方法:

func followIconTapped(sender: UITapGestureRecognizer) {
    ...
}

而且一切正常。但我需要将 channelId 传递给 followIconTapped() 方法。

我试试这个:

func followUnfollow(followIcon: UIImageView, channelId: String) {
    ...
    let tapGestureRecognizer = UITapGestureRecognizer(target: self, action: #selector(self.followIconTapped(_:channelId)))
    ...
}

然后我试着捕获它:

func followIconTapped(sender: UITapGestureRecognizer, channelId: String) {
    ...
}

xCode 说 channelId 永远不会被使用。为什么？ 当我构建项目时，我没有任何问题。但是，如果我点击 followIcon，应用程序会崩溃。

请给我一些建议，如何将 channelId 传递给 followIconTapped()

Answer 1

创建一个通用的 UITapGestureRecognizer 而不是使用这个:

class CustomTapGestureRecognizer: UITapGestureRecognizer {
    var channelId: String?
}

也用这个:

override func viewDidLoad() {
    super.viewDidLoad()

    let gestureRecognizer = CustomTapGestureRecognizer(target: self, action: #selector(tapped(_:))
    gestureRecognizer.channelId = "Your string"
    view1.addGestureRecognizer(gestureRecognizer)
}

func tapped(gestureRecognizer: CustomTapGestureRecognizer) {
    if let channelId = gestureRecognizer.channelId {
        //print
    }
}