Sign Up

Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.

Have an account? Sign In

Have an account? Sign In Now

Sign In

Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.

Sign Up Here

Forgot Password?

Don't have account, Sign Up Here

Forgot Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

Have an account? Sign In Now

You must login to ask a question.

Forgot Password?

Need An Account, Sign Up Here

Please briefly explain why you feel this question should be reported.

Please briefly explain why you feel this answer should be reported.

Please briefly explain why you feel this user should be reported.

Sign InSign Up

The Archive Base

The Archive Base Logo The Archive Base Logo

The Archive Base Navigation

  • SEARCH
  • Home
  • About Us
  • Blog
  • Contact Us
Search
Ask A Question

Mobile menu

Close
Ask a Question
  • Home
  • Add group
  • Groups page
  • Feed
  • User Profile
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Buy Points
  • Users
  • Help
  • Buy Theme
  • SEARCH
Home/ Questions/Q 6152699
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: May 23, 20262026-05-23T19:57:54+00:00 2026-05-23T19:57:54+00:00

So I am constantly adding new points to a graph and I want all

  • 0

So I am constantly adding new points to a graph and I want all of them to have the same click function. But it seems like when I run

$('.someClass').click(function(){})

it only applies to elements that currently have someClass. If I add a new someClass element, it does not have the click listener.

How do I get around this? Must I run click function every time I add an element?

  • 1 1 Answer
  • 0 Views
  • 0 Followers
  • 0
Share
  • Facebook
  • Report

Leave an answer
Cancel reply

You must login to add an answer.

Forgot Password?

Need An Account, Sign Up Here

1 Answer

  • Voted
  • Oldest
  • Recent
  • Random
  1. Editorial Team
    Editorial Team
    2026-05-23T19:57:55+00:00Added an answer on May 23, 2026 at 7:57 pm
    $('.someClass').live('click', function(){});
    

    Or even better consider delegate method.

    • 0
    • Reply
    • Share
      Share
      • Share on Facebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
      • Report

Sidebar

Related Questions

I have a FlowLayoutPanel, and I am constantly adding controls to it. They all
I am constantly learning new tools, even old fashioned ones, because I like to
My coworker and I constantly argue about button sizes. I like to have buttons
I have folder that I would like to include in my Xcode project, but
I have a script that constantly segfaults - the problem that I can't solve
I have a command line Ruby app I'm developing and I want to allow
I'm creating locally a big database using MySQL and PHPmyAdmin. I'm constantly adding a
I am new to Visual Studio C#. I receive the following error constantly when
Here is my jQuery using delegate and ajax: $(.tweets).delegate(#fav #submit, click, function(){ var favid
I have a chart created as shown below, and I am adding values to

Explore

  • Home
  • Add group
  • Groups page
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Users
  • Help
  • SEARCH

Footer

© 2021 The Archive Base. All Rights Reserved
With Love by The Archive Base

Insert/edit link

Enter the destination URL

Or link to existing content

    No search term specified. Showing recent items. Search or use up and down arrow keys to select an item.