我有一个UIView并且我添加了点击手势

let tap = UITapGestureRecognizer(target: self, action: Selector("handleTap:"))
tap.delegate = self
myView.addGesture(tap)

我试图在testfile中以编程方式调用它。

sendActionForEvent

我正在使用这个函数,但它不起作用:

myView.sendActionForEvent(UIEvents.touchUpDown)

它显示未识别的选择器发送到实例。

我该如何解决这个问题呢?


当前回答

Swift 5.1三视图示例

步骤:1 ->添加storyboard视图和outlet viewController UIView

@IBOutlet var firstView: UIView!
@IBOutlet var secondView: UIView!
@IBOutlet var thirdView: UIView!

步骤:2 ->添加storyBoard视图标签

步骤:3 ->添加手势

override func viewDidLoad() {
        super.viewDidLoad()

        firstView.addGestureRecognizer(UITapGestureRecognizer(target: self, action: #selector(self.tap(_:))))
        firstView.isUserInteractionEnabled = true
        secondView.addGestureRecognizer(UITapGestureRecognizer(target: self, action: #selector(self.tap(_:))))
        secondView.isUserInteractionEnabled = true
        thirdView.addGestureRecognizer(UITapGestureRecognizer(target: self, action: #selector(self.tap(_:))))
        thirdView.isUserInteractionEnabled = true
    }

步骤:4 -> select view

@objc func tap(_ gestureRecognizer: UITapGestureRecognizer) {
        let tag = gestureRecognizer.view?.tag
        switch tag! {
        case 1 :
            print("select first view")
        case 2 :
            print("select second view")
        case 3 :
            print("select third view")
        default:
            print("default")
        }
    }

其他回答

你需要用目标和动作初始化UITapGestureRecognizer,如下所示:

let tap = UITapGestureRecognizer(target: self, action: "handleTap:")
tap.delegate = self
myView.addGestureRecognizer(tap)

然后,你应该实现处理程序,它将在每次tap事件发生时被调用:

func handleTap(sender: UITapGestureRecognizer) {
  // handling code
}

对于Swift 4:

let tap = UITapGestureRecognizer(target: self, action: #selector(self.handleTap(_:)))

view.addGestureRecognizer(tap)

view.isUserInteractionEnabled = true

self.view.addSubview(view)

// function which is triggered when handleTap is called
@objc func handleTap(_ sender: UITapGestureRecognizer) {
    print("Hello World")
}

在Swift 4中,你需要显式地指出触发的函数是可从Objective-C调用的,所以你需要添加@objc你的handleTap函数。

请看@Ali Beadle的回答:Swift 4添加手势:覆盖vs @objc

试试下面的扩展

extension UIView {
    
    func  addTapGesture(action : @escaping ()->Void ){
        let tap = MyTapGestureRecognizer(target: self , action: #selector(self.handleTap(_:)))
        tap.action = action
        tap.numberOfTapsRequired = 1
        
        self.addGestureRecognizer(tap)
        self.isUserInteractionEnabled = true
        
    }

    @objc func handleTap(_ sender: MyTapGestureRecognizer) {
        sender.action!()
    }
}

class MyTapGestureRecognizer: UITapGestureRecognizer {
    var action : (()->Void)? = nil
}

然后使用它:

submitBtn.addTapGesture {
     //your code
}

你甚至可以把它用于cell

cell.addTapGesture {
     //your code
}

1 .

@IBOutlet var viewTap: UIView!

2 .

var tapGesture = UITapGestureRecognizer()

3 .

override func viewDidLoad() {
    super.viewDidLoad()
    // TAP Gesture
    tapGesture = UITapGestureRecognizer(target: self, action: #selector(ViewController.myviewTapped(_:)))
    tapGesture.numberOfTapsRequired = 1
    tapGesture.numberOfTouchesRequired = 1
    viewTap.addGestureRecognizer(tapGesture)
    viewTap.isUserInteractionEnabled = true
}

4 .

func myviewTapped(_ sender: UITapGestureRecognizer) {

    if self.viewTap.backgroundColor == UIColor.yellow {
        self.viewTap.backgroundColor = UIColor.green
    }else{
        self.viewTap.backgroundColor = UIColor.yellow
    }
}

输出

xCode 9.3, Swift 4.0

class BaseVC: UIViewController, UIGestureRecognizerDelegate { 

      @IBOutlet weak var iView: UIView!

      override func viewDidLoad() {
          super.viewDidLoad()
          let clickUITapGestureRecognizer = UITapGestureRecognizer(target: self, action: #selector(self.onSelect(_:)))
          clickUITapGestureRecognizer.delegate = self
          iView?.addGestureRecognizer(tap)
      }

      func gestureRecognizer(_ gestureRecognizer: UIGestureRecognizer, shouldReceive touch: UITouch) -> Bool {
          return true
      }


     @IBAction func onSelect(_ sender: Any) {

     }
}