'UITableViewDiffableDataSource is not calling tableView(_:commit:forRowAt:)
There is a UITableView that uses UITableViewDiffableDataSource. I subclassed UITableViewDiffableDataSource to add canEditRowAt. This correctly shows the swipeable delete action when gesturing the row. However, clicking the delete option does not call tableView(_:commit:forRowAt:). I have read that you have to use tableView(_:trailingSwipeActionsConfigurationForRowAt:indexPath:) because the other function is not supported. I wanted to confirm that was true. If we subclass tableView(_:commit:forRowAt:) as well, we need a clean way to call a function on the original View Controller.
// MyViewController
override func tableView(_ tableView: UITableView, commit editingStyle: UITableViewCell.EditingStyle, forRowAt indexPath: IndexPath) {
if editingStyle == .delete {
// Delete cell
}
}
// Subclass DiffableDataSource used in MyViewController
final class CustomDiffableDatasource: UITableViewDiffableDataSource<MyViewController.Section, MyViewController.Item> {
override func tableView(_ tableView: UITableView, canEditRowAt indexPath: IndexPath) -> Bool {
guard let item = itemIdentifier(for: indexPath) else {
return false
}
return item.isEditable
}
}
Solution 1:[1]
Here is the custom trailingSwipeActionsConfigurationForRowAtfunc looks like for adding a delete swipe. I saw this in another StackOverflow question that referenced a blog post.
override func tableView(_ tableView: UITableView, trailingSwipeActionsConfigurationForRowAt indexPath: IndexPath) -> UISwipeActionsConfiguration? {
guard self.dataSource?.tableView(tableView, canEditRowAt: indexPath) == true else {
return nil
}
let delete = UIContextualAction(style: .destructive, title: "Delete") { [weak self] action, view, success in
self?.remove(at: indexPath)
success(true)
}
let swipeActionConfig = UISwipeActionsConfiguration(actions: [delete])
swipeActionConfig.performsFirstActionWithFullSwipe = false
return swipeActionConfig
}
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | Josh Birdwell |
