How to identify a clickable Stack View in UIKit?-CodePudding

I am building an app using UIKit programmatically. I have a for loop to render 5 UIStackViews on the screen, every StackView has a click gesture that triggers a function.

 private func configureCardsMainContainer() {
        cardsMainContainer = UIStackView()
       //...
        for i in 0..<5 {
            configureCardsHContainer(index: i)
        }
        //...
    }

   private func configureCardsHContainer(index:Int) {
        cardsHContainer = UIStackView()
        //...
        let tapGesture = UITapGestureRecognizer(target: self, action: #selector(clickView(_:)))
        tapGesture.delegate = self
        cardsHContainer.addGestureRecognizer(tapGesture)
    }

    @objc func clickView(_ sender: UIStackView) {
        let details = DetailsViewController()
        self.navigationController?.pushViewController(details, animated: true)
    }

After clicking a clickable StackView it navigates to DetailsViewController, I want to set the navigation title of DetailsViewController to be something different for each StackView that has been clicked. The problem is, I cannot identify the sender view, is there any way to identify a sender which is a StackView?

CodePudding user response：

Simple, put your stacks into an array.

var stacks: [UIStackView] = []

private func configureCardsMainContainer() {
   cardsMainContainer = UIStackView()
   //...

   stacks = 0..<5.map { createCardsHContainer(index: $0) }

   //...
}

private func createCardsHContainer(index:Int) -> UIStackView {
    let cardsHContainer = UIStackView()
    //...
    let tapGesture = UITapGestureRecognizer(target: self, action: #selector(clickView(_:)))
    tapGesture.delegate = self
    cardsHContainer.addGestureRecognizer(tapGesture)
    
    return cardsHContainer
}

@objc private func clickView(_ sender: UIStackView) {
    guard let index = stacks.firstIndex(where: { $0 === sender }) else { return }

    ...
}

Of course, if you extract your repetitive component into a separate view, this can be done even simpler.