ios - 如何在静态函数中使用 UITapGestureRecognizer 向标签添加点击功能

Question

我需要在我的应用程序中动态地向 labels 添加点击功能(以打开网站)，我想在 class 中创建静态函数。我想在我的应用程序的任何 ViewController 中启动该功能。

我使用 Swift 3 完成了这个:

class Launcher {
    static func openWeb(label: UILabel, url: String) {
        func tapFunction(sender:UITapGestureRecognizer) {

            if #available(iOS 10.0, *) {
                UIApplication.shared.open(URL(string: url)!, options: [:], completionHandler: nil)
            } else {
                UIApplication.shared.openURL(URL(string: url)!)
            }
        }

        let tap = UITapGestureRecognizer(target: self, action: #selector(tapFunction))
        label.isUserInteractionEnabled = true
        label.addGestureRecognizer(tap)

        if #available(iOS 10.0, *) {
            UIApplication.shared.open(URL(string: url)!, options: [:], completionHandler: nil)
        } else {
            UIApplication.shared.openURL(URL(string: url)!)
        }

    }
}

但不起作用，因为我在 action: #selector(tapFunction)

中遇到错误

错误:“#selector”的参数不能引用本地函数“tapFunction(sender:)”

如果我使用 action: #selector(myView.tapFunction) 在 ViewController 代码中使用它，如下所示，有效

viewDidLoad 内部

let tap = UITapGestureRecognizer(target: self, action: #selector(MyViewController.tapFunction))
mylabel.isUserInteractionEnabled = true
mylabel.addGestureRecognizer(tap)

ViewController 中的分离函数

func tapFunction(sender:UITapGestureRecognizer) {

        if #available(iOS 10.0, *) {
            UIApplication.shared.open(URL(string: url)!, options: [:], completionHandler: nil)
        } else {
            UIApplication.shared.openURL(URL(string: url)!)
        }
    }

我想将这最后一段代码转换为我的 Class 中的静态函数，以便在任何 ViewController 中调用它。谢谢

Answer 1

Try resolve it using extension

extension UILabel {
    func openWeb(url: String) {
        let tap = UITapGestureRecognizer(target: self, action: #selector(tapFunction))
        self.isUserInteractionEnabled = true
        self.addGestureRecognizer(tap)

        openURL(url: url)
    }

    func tapFunction(sender:UITapGestureRecognizer) {
        openURL(url: "")
    }

    func openURL(url: String) {
        if #available(iOS 10.0, *) {
            UIApplication.shared.open(URL(string: url)!, options: [:], completionHandler: nil)
        } else {
            UIApplication.shared.openURL(URL(string: url)!)
        }
    }
}

let label = UILabel()
label.openWeb(url: "123")

To store link into label you can use associations with label object.