带有参数 gestureRecognizer 的选择器

Question

Argument of '#selector' does not refer to an '@objc' method, property, or initializer

问题：当我尝试使用选择器传递参数时出现以上错误代码片段：

let gestureRecognizer = UITapGestureRecognizer(target: self, action: #selector(labelPressed(i: 1)))

func labelPressed(i: Int){
        print(i)
}

Answer 1

你应该像这样声明函数：

@objc func labelPressed(i: Int){ print(i) }

Answer 2

Swift3 的更新：

使用更现代的语法，您可以这样声明您的函数：

@objc func labelTicked(withSender sender: AnyObject) {

并像这样初始化你的手势识别器，使用#selector:

UITapGestureRecognizer(target: self, action: #selector(labelTicked(withSender:)))

Answer 3

您不能将参数传递给这样的函数。动作 - 这是只传递发送者，在这种情况下是手势识别器。您想要做的是获取您将手势附加到的 UIView：

let gestureRecognizer = UITapGestureRecognizer(target: self, action: #selector(labelPressed())

func labelPressed(_ recognizer:UITapGestureRecognizer){
    let viewTapped = recognizer.view
}

更多注意事项：

(1) You may only attach a single view to a recognizer.
(2) You might want to use both the `tag` property along with the `hitTest()` method to know which subview was hit. For example:


let view1 = UIView()
let view2 = UIView()

// add code to place things, probably using auto layout

view1.tag = 1
view2.tag = 2
mainView.addSubview(view1)
mainView.addSubview(view2)

let gestureRecognizer = UITapGestureRecognizer(target: self, action: #selector(mainViewTapped())

func mainViewTapped(_ recognizer:UITapGestureRecognizer){

    // get the CGPoint of the tap

    let p = recognizer.location(in: self)
    let viewTapped:UIView!

    // there are many (better) ways to do this, but this works

    for view in self.subviews as [UIView] {
        if view.layer.hitTest(p) != nil {
            viewTapped = view
        }
    }

    // if viewTapped != nil, you have your subview

}

带有参数 gestureRecognizer 的选择器

selector with argument gestureRecognizer

xcode

uigesturerecognizer

swift