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Home/ Questions/Q 842513
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Editorial Team
  • 0
Editorial Team
Asked: May 15, 20262026-05-15T05:58:31+00:00 2026-05-15T05:58:31+00:00

This isn’t pretty, but I think you’ll get what I’m trying to do. <label><input

  • 0

This isn’t pretty, but I think you’ll get what I’m trying to do.

<label><input name="jpgSel" type="radio" value="0">jpg 1</label><br />
<label><input name="jpgSel" type="radio" value="1">jpg 2</label><br />
<div id="showjpg"></div>

and

$(document).ready(function() {
    $("select").change(function () {
           if ($("jpgSel:checked").val() == 1) {
            $('#showjpg').attr({
                src: 'one.jpg',
                 alt: 'one dot jpg'
        });
    }
    else {
            $('#showjpg').attr({
                 src: 'two.jpg',
                 alt: 'two dot jpg'
        });
    }
});
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-15T05:58:32+00:00Added an answer on May 15, 2026 at 5:58 am

    You have several issues:

    • The #showjpg element is a div, you should have there an img element.
    • The selector you are using to bind the change event is incorrect ($("select")), it should be something like $("input[name=jpgSel]")
    • If you bind the change event to both radio buttons, you can get the value of the selected element simply by this.value.

    Try this:

    $(document).ready(function() {
      $("input[name=jpgSel]").change(function () {
        if (this.value == "0") {
          $('#showjpg').attr({
            src: 'one.jpg',
            alt: 'one dot jpg'
          });
        } else {
          $('#showjpg').attr({
            src: 'two.jpg',
            alt: 'two dot jpg'
          });
        }
      });
    });
    

    Check an example here.

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