带参数的选择器gestureRecognizer

Posted 2023-03-13

【中文标题】带参数的选择器gestureRecognizer

【英文标题】：selector with argument gestureRecognizer

【发布时间】：2017-05-20 04:47:57

【问题描述】：

“#selector”的参数不引用“@objc”方法、属性、 或初始化器

问题：当我尝试使用选择器传递参数时出现上述错误 代码sn-p：

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

func labelPressed(i: Int)
        print(i)

【问题讨论】：

Selector to get indexPath UICollectionView Swift 3.0的可能重复

【参考方案1】：

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

还有一些注意事项：

(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

【讨论】：

你知道如何为点击设置动画吗？ Like按钮，当颜色改变时禁食。

【参考方案2】：

你应该像这样声明函数：

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

【讨论】：

这行得通吗？手势是发送者，而不是 UILabel。

【参考方案3】：

Swift 3 更新：

使用更现代的语法，您可以像这样声明您的函数：

@objc func labelTicked(withSender sender: AnyObject) 

然后像这样初始化你的手势识别器，使用#selector:

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