Swift：在 table 内单击以编程方式生成的视图

Question

尝试了很多东西，但 none 成功了。

我当前的视图层次结构：

TableView 单元格
垂直堆栈视图

以编程方式生成的视图 - 未触发手势识别器

let vContainer = UIView()
vContainer.translatesAutoresizingMaskIntoConstraints = false
vContainer.isUserInteractionEnabled = true
vContainer.backgroundColor = .red
//Some view generation inside box is skipped (can be seen on image)
self.vsvViewContainer.addArrangedSubview(vContainer)
var tapRecognizer = UITapGestureRecognizer(target: self, action: #selector(showValuePicker(sender:)))
vContainer.addGestureRecognizer(tapRecognizer)

稍后 class：

    @objc private func showValuePicker(sender: UITapGestureRecognizer) {
        print("showValuePicker")
    }

红框是View，我要添加点击的地方

Answer 1

将 tapRecognizer.cancelsTouchesInView = false 添加到您的代码中

self.vsvViewContainer.addArrangedSubview(vContainer)
var tapRecognizer = UITapGestureRecognizer(target: self, action: #selector(showValuePicker(sender:)))
tapRecognizer.cancelsTouchesInView = false
vContainer.addGestureRecognizer(tapRecognizer)

Answer 2

为什么指定目标为 nil？

在：

var tapRecognizer = UITapGestureRecognizer(target: nil, action: #selector(showValuePicker(sender:)))

来自documentation：

Target

An object that is the recipient of action messages sent by the receiver when it recognizes a gesture. nil is not a valid value.

您可以尝试添加 self 作为目标吗？

Answer 3

发现问题...

"self" 指向视图生成 class，与视图控制器

无关

将点击事件直接传递给视图控制器解决了问题

Swift：在 table 内单击以编程方式生成的视图

Swift: click inside programmatically generated view inside table

uitapgesturerecognizer

swift