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Home/ Questions/Q 6606977
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Editorial Team
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Editorial Team
Asked: May 25, 20262026-05-25T19:27:12+00:00 2026-05-25T19:27:12+00:00

How do I go about doing this? $(function() { var foo = $(‘#foo’), bar

  • 0

How do I go about doing this?

$(function() {

   var foo = $('#foo'),
       bar = $('#bar');

    $('body').click(function() {

        $(foo,bar).css({color: 'red'}); 

    });

});

Demo: http://jsfiddle.net/each/RGZ4Z/ – Only foo becomes red

Edit: Can i stress the fact that I’m aware I could easily do:

$('#foo,#bar').css({color: 'red'});

I’m simply asking about the usage of variables…

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-25T19:27:13+00:00Added an answer on May 25, 2026 at 7:27 pm

    Probably, what you want is this:

    foo.add(bar).css({color: 'red'});
    

    Here’s a demo piece of code that shows how this can work: http://jsfiddle.net/jfriend00/3Ep6J/.

    There are three choices I can think of depending upon what you want:

    1. var items = $('#foo, #bar')
    2. $('#foo, #bar').css({color: 'red'});
    3. foo.add(bar).css({color: 'red'});

    And, more detail on each option:

    1) You can either just create one jQuery object in the beginning that has both sets of objects in it like this:

    var objs = $('#foo, #bar');
    

    and then later do this:

    objs.css({color: 'red'}); 
    

    2) Or just do it all at once:

    $('#foo, #bar').css({color: 'red'}); 
    

    3) Or, if you already have separate jQuery objects for foo and bar, you can add the items from one jQuery object to the other and then carry out your operation:

    var foo = $('#foo'),
        bar = $('#bar');
    

    and then later, do this:

    foo.add(bar).css({color: 'red'}); 
    

    Note: somewhat counterintuively, option 3) does not modify the foo jQuery object, the add method returns a new jQuery object with the items from bar added to it.

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