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

Swift 4

This is best way to get indexPath using touchPoint

class YourTableViewController: UITableViewController {

    // ...

    override func tableView(_ tableView: UITableView, cellForRowAt indexPath: IndexPath) -> UITableViewCell {
        let cell = tableView.dequeueReusableCell(withIdentifier: "SwiftyCell", for: indexPath) as! SwiftyTableViewCell

        cell.label.text = "This is cell number \(indexPath.row)"

        // WRONG! When cells get reused, these actions will get added again! That's not what we want.
        // Of course, we could get around this by jumping through some hoops, but maybe there's a better solution...
        cell.yourButton.addTarget(self, action: #selector(self.yourButtonTapped(_:)), for: .touchUpInside)

        return cell
    }

    func yourButtonTapped(_ sender: Any?) {
        let point = tableView.convert(sender.center, from: sender.superview!)

        if let wantedIndexPath = tableView.indexPathForItem(at: point) {
            let cell = tableView.cellForItem(at: wantedIndexPath) as! SwiftyCell

        }
    }
    // ...

}

For more details you can follow this tutorials