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 571233
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: May 13, 20262026-05-13T13:29:08+00:00 2026-05-13T13:29:08+00:00

I need to apply this script to work the same way for two other

  • 0

I need to apply this script to work the same way for two other divs:
This works well, but how can I make it do the same thing for two other elements without rewriting the entire thing for each?

$(".survey_option li > a").each().live('click', function (event) {
    // First disable the normal link click
    event.preventDefault();

    // Remove all list and links active class.
    $('.so_1 .active').toggleClass("active");
    // Grab the link clicks ID
    var id = this.id;

    // The id will now be something like "link1"
    // Now we need to replace link with option (this is the ID's of the checkbox)
    var newselect = id.replace('partc', 'optionc');

    // Make newselect the option selected.
    $('#'+newselect).attr('checked', true);

    // Now add active state to the link and list item
    $(this).addClass("active").parent().addClass("active");

    return false;
});
  • 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
    2026-05-13T13:29:08+00:00Added an answer on May 13, 2026 at 1:29 pm

    You can cause it to affect multiple elements by changing your selector:

    $(".element_1, .element_2, .randomElement, #someOtherThing");
    

    Note also that you don’t need to call $.each() before adding your $.live() call. $.live() will be applied to all of the matched elements, doing its own internal $.each() logic.

    $(".elem1, .elem2 > a, #someThing").live("click", function(e){
      e.preventDefault();
      alert($(this).text()); // works on all elements in the selector
    });
    
    • 0
    • Reply
    • Share
      Share
      • Share on Facebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
      • Report

Sidebar

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.