使用UITapGestureRecognizer传递参数

Question

有什么办法可以用UITapGestureRecognizer传递参数吗？我已经看到这回答了Objective-c但是找不到快速的答案

test.userInteractionEnabled = true
let tapRecognizer = UITapGestureRecognizer(target: self, action: Selector("imageTapped4:"))
// Something like text.myParamater
test.addGestureRecognizer(tapRecognizer)

然后在func imageTapped4(){}下接收myParameter

Answer 1

一种方法是将UITapGestureRecognizer子类化,然后设置属性,我在下面发布了一个例子.您还可以检查发件人并检查是否等于某些标签,类,字符串等

class ViewController: UIViewController {

    @IBOutlet weak var label1: UILabel!
    @IBOutlet weak var image: UIImageView!

    override func viewDidLoad() {
        super.viewDidLoad()
        // Do any additional setup after loading the view, typically from a nib.
        image.userInteractionEnabled = true;
        let tappy = MyTapGesture(target: self, action: #selector(self.tapped(_:)))
        image.addGestureRecognizer(tappy)
        tappy.title = "val"
    }

    func tapped(sender : MyTapGesture) {
        print(sender.title)
        label1.text = sender.title
    }
}

class MyTapGesture: UITapGestureRecognizer {
    var title = String()
}

SO上有很多例子,看看,祝你好运.

Answer 2

对于Swift 4

在Viewdidload中

let label     =   UILabel(frame: CGRect(x: 0, y: h, width: Int(self.phoneNumberView.bounds.width), height: 30))
                label.textColor = primaryColor
                label.numberOfLines = 0
                label.font = title3Font
                label.lineBreakMode = .byWordWrapping
                label.attributedText = fullString

 let phoneCall = MyTapGesture(target: self, action: #selector(self.openCall))
        phoneCall.phoneNumber = "\(res)"
            label.isUserInteractionEnabled = true
            label.addGestureRecognizer(phoneCall)

功能为

@objc func openCall(sender : MyTapGesture) {
        let number = sender.phoneNumber
        print(number)
}

写类为

class MyTapGesture: UITapGestureRecognizer {
    var phoneNumber = String()
}

正确执行步骤，然后根据您的变量，按钮，标签进行更改。它可以正常工作