Conversation
If you've ever used .clone() in jQuery, you have encountered an issue where buttons become unresponsive after cloning. This occurs because the cloned element is already rendered, and that's what we need to address. To resolve this, we've implemented a check to determine if an element has already been rendered. If it has, we remove all elements added in the previous call to the .render() function and then proceed to render it once more.
This file contains hidden or bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
Sign up for free
to join this conversation on GitHub.
Already have an account?
Sign in to comment
1 participant
Add this suggestion to a batch that can be applied as a single commit.This suggestion is invalid because no changes were made to the code.Suggestions cannot be applied while the pull request is closed.Suggestions cannot be applied while viewing a subset of changes.Only one suggestion per line can be applied in a batch.Add this suggestion to a batch that can be applied as a single commit.Applying suggestions on deleted lines is not supported.You must change the existing code in this line in order to create a valid suggestion.Outdated suggestions cannot be applied.This suggestion has been applied or marked resolved.Suggestions cannot be applied from pending reviews.Suggestions cannot be applied on multi-line comments.Suggestions cannot be applied while the pull request is queued to merge.Suggestion cannot be applied right now. Please check back later.
If you've ever used .clone() in jQuery, you have encountered an issue where buttons become unresponsive after cloning. This occurs because the cloned element is already rendered, and that's what we need to address. To resolve this, we've implemented a check to determine if an element has already been rendered. If it has, we remove all elements added in the previous call to the .render() function and then proceed to render it once more.