Open URL with a button inside a table view cell

Question

I want to include a button in each table cell that opens a URL.

I\'ve created tables (using an array) with images and labels just fine, however I\'m confused how to crea

Answer 1

Here's how I usually solve this. Create a delegate for your UITableViewCell subclass, and set the view controller owning the tableView as its delegate. Add methods for the interactions that happens inside the cell.

protocol YourTableViewCellDelegate: class {
    func customCellDidPressUrlButton(_ yourTableCell: YourTableViewCell)
}

class YourTableViewCell: UITableViewCell {
    weak var delegate: YourTableViewCellDelegate?

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

        let button = UIButton()
        button.addTarget(self, action: #selector(buttonTapped), for: .touchUpInside)
        addSubview(button)
    }

    required init?(coder _: NSCoder) {
        return nil
    }

    @objc func buttonTapped() {
        delegate?.customCellDidPressUrlButton(self)
    }

}

Then, in the controller, set itself as a delegate and get the indexPath trough the proper method, indexPath(for:)

class YourTableViewController: UITableViewController {

    override func tableView(_ tableView: UITableView, cellForRowAt indexPath: IndexPath) -> UITableViewCell {
        let cell = tableView.dequeueReusableCell(withIdentifier: "cell", for: indexPath) as! YourTableViewCell
        cell.delegate = self
        return cell
    }
}

extension YourTableViewController: YourTableViewCellDelegate {
    func customCellDidPressUrlButton(_ yourTableCell: YourTableViewCell) {
        guard let indexPath = tableView.indexPath(for: yourTableCell) else { return }
        print("Link button pressed at \(indexPath)")
    }
}

Then use that indexPath to grab the correct URL and present it from your table viewcontroller with a SFSafariViewController.