UIMenu from Button Inside UITableView Not Showing Properly? Fix It Like a Pro!

Are you stuck with a frustrating issue where your UIMenu from a button inside a UITableView cell is not showing up as expected? Don’t worry, you’re not alone! Many developers have faced this problem, and in this article, we’ll dive deep into the solutions to get your context menu up and running in no time.

Understanding the Problem

The issue arises when you try to present a UIMenu from a button inside a UITableViewCell. The menu might not appear at all, or it might appear but not in the correct position. This can be attributed to the way UITableViewCell handles touch events and the layout of its subviews.

Why Does This Happen?

Here are some possible reasons why your UIMenu might not be showing properly:

• Cell Reuse Issue: When you reuse table view cells, the button’s superview might not be the cell itself, but rather the table view. This can cause the menu to appear at an incorrect position or not at all.
• Subview Hierarchy: The button’s position in the subview hierarchy can affect the menu’s presentation. If the button is not a direct subview of the cell’s content view, the menu might not appear correctly.
• Touch Event Handling: UITableView cells can intercept touch events, which might prevent the button from receiving the necessary events to present the menu.

Solutions to Make It Work

Now that we’ve identified the possible causes, let’s explore the solutions to overcome this issue:

Solution 1: Use a Custom UITableViewCell

By creating a custom UITableViewCell class, you can ensure that the button is a direct subview of the cell’s content view. This can help resolve the subview hierarchy issue.

// CustomTableViewCell.swift
import UIKit

class CustomTableViewCell: UITableViewCell {
    let button = UIButton(type: .system)

    override init(style: UITableViewCell.CellStyle, reuseIdentifier: String?) {
        super.init(style: style, reuseIdentifier: reuseIdentifier)
        setupButton()
    }

    required init?(coder: NSCoder) {
        fatalError("init(coder:) has not been implemented")
    }

    func setupButton() {
        button.setTitle("Show Menu", for: .normal)
        button.addTarget(self, action: #selector(showMenu), for: .touchUpInside)
        contentView.addSubview(button)
        button.translatesAutoresizingMaskIntoConstraints = false
        button.centerXAnchor.constraint(equalTo: contentView.centerXAnchor).isActive = true
        button.centerYAnchor.constraint(equalTo: contentView.centerYAnchor).isActive = true
    }

    @objc func showMenu() {
        // Create and present the menu
        let menu = UIMenu(title: "My Menu", children: [
            UIAction(title: "Action 1", handler: { _ in
                print("Action 1 tapped")
            }),
            UIAction(title: "Action 2", handler: { _ in
                print("Action 2 tapped")
            }),
        ])
        button.showsMenuAsPrimaryAction = true
        button.menu = menu
    }
}

Solution 2: Use a UITapGestureRecognizer

Another approach is to add a UITapGestureRecognizer to the UITableViewCell’s content view. This can help bypass the cell’s touch event handling and allow the button to receive the necessary events.

// ViewController.swift
import UIKit

class ViewController: UIViewController, UITableViewDataSource, UITableViewDelegate {
    @IBOutlet weak var tableView: UITableView!

    override func viewDidLoad() {
        super.viewDidLoad()
        tableView.delegate = self
        tableView.dataSource = self
    }

    func tableView(_ tableView: UITableView, cellForRowAt indexPath: IndexPath) -> UITableViewCell {
        let cell = tableView.dequeueReusableCell(withIdentifier: "Cell", for: indexPath)
        let button = UIButton(type: .system)
        button.setTitle("Show Menu", for: .normal)
        cell.contentView.addSubview(button)
        // Add gesture recognizer
        let tapGestureRecognizer = UITapGestureRecognizer(target: self, action: #selector(showMenu(_:)))
        button.addGestureRecognizer(tapGestureRecognizer)
        return cell
    }

    @objc func showMenu(_ sender: UITapGestureRecognizer) {
        // Create and present the menu
        let menu = UIMenu(title: "My Menu", children: [
            UIAction(title: "Action 1", handler: { _ in
                print("Action 1 tapped")
            }),
            UIAction(title: "Action 2", handler: { _ in
                print("Action 2 tapped")
            }),
        ])
        if let button = sender.view as? UIButton {
            button.showsMenuAsPrimaryAction = true
            button.menu = menu
        }
    }
}

Solution 3: Disable Cell Selection and User Interaction

You can also disable cell selection and user interaction to prevent the cell from intercepting touch events. This might not be suitable for all scenarios, but it can be a viable solution in some cases.

// ViewController.swift
import UIKit

class ViewController: UIViewController, UITableViewDataSource, UITableViewDelegate {
    @IBOutlet weak var tableView: UITableView!

    override func viewDidLoad() {
        super.viewDidLoad()
        tableView.delegate = self
        tableView.dataSource = self
    }

    func tableView(_ tableView: UITableView, cellForRowAt indexPath: IndexPath) -> UITableViewCell {
        let cell = tableView.dequeueReusableCell(withIdentifier: "Cell", for: indexPath)
        let button = UIButton(type: .system)
        button.setTitle("Show Menu", for: .normal)
        cell.contentView.addSubview(button)
        // Disable cell selection and user interaction
        cell.selectionStyle = .none
        cell.isUserInteractionEnabled = false
        button.addTarget(self, action: #selector(showMenu(_:)), for: .touchUpInside)
        return cell
    }

    @objc func showMenu(_ sender: UIButton) {
        // Create and present the menu
        let menu = UIMenu(title: "My Menu", children: [
            UIAction(title: "Action 1", handler: { _ in
                print("Action 1 tapped")
            }),
            UIAction(title: "Action 2", handler: { _ in
                print("Action 2 tapped")
            }),
        ])
        sender.showsMenuAsPrimaryAction = true
        sender.menu = menu
    }
}

Bonus Solution: Using a UIControl as a Container

In some cases, you might want to use a UIControl as a container to receive touch events and present the menu. This can be useful when you have a complex layout or multiple buttons in a cell.

// ViewController.swift
import UIKit

class ViewController: UIViewController, UITableViewDataSource, UITableViewDelegate {
    @IBOutlet weak var tableView: UITableView!

    override func viewDidLoad() {
        super.viewDidLoad()
        tableView.delegate = self
        tableView.dataSource = self
    }

    func tableView(_ tableView: UITableView, cellForRowAt indexPath: IndexPath) -> UITableViewCell {
        let cell = tableView.dequeueReusableCell(withIdentifier: "Cell", for: indexPath)
        let container = UIControl()
        cell.contentView.addSubview(container)
        container.translatesAutoresizingMaskIntoConstraints = false
        container.topAnchor.constraint(equalTo: cell.contentView.topAnchor).isActive = true
        container.bottomAnchor.constraint(equalTo: cell.contentView.bottomAnchor).isActive = true
        container.leadingAnchor.constraint(equalTo: cell.contentView.leadingAnchor).isActive = true
        container.trailingAnchor.constraint(equalTo: cell.contentView.trailingAnchor).isActive = true
        let button = UIButton(type: .system)
        button.setTitle("Show Menu", for: .normal)
        container.addSubview(button)
        button.translatesAutoresizingMaskIntoConstraints = false
        button.centerXAnchor.constraint(equalTo: container.centerXAnchor).isActive = true
        button.centerYAnchor.constraint(equalTo: container.centerYAnchor).isActive = true
        container.addTarget(self, action: #selector(showMenu(_:)), for: .touchUpInside)
        return cell
    }

    @objc func showMenu(_ sender: UIControl) {
        // Create and present the menu
        let menu = UIMenu(title: "My Menu", children: [
            UIAction(title: "Action 1", handler: { _ in
                print("Action 1 tapped")
            }),
            UIAction(title: "Action 2", handler: { _ in
                print("Action 2 tapped")
            }),
        ])
        if let button = sender.subviews.first as? UIButton {
            button.showsMenuAsPrimaryAction = true
            button.menu = menu
        }
    }
}

Conclusion

In this article, we’ve explored three solutions to overcome the issue of UIMenu not showing properly when presented from a button inside a UITableView cell. By using a custom UITableViewCell, adding a UITapGestureRecognizer, or disabling cell selection and user interaction, you can ensure that your context menu appears correctly. Remember to adapt these solutions to your specific use case, and don’t hesitate to experiment with different approaches until you find the one that works best for your app.