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Home/ Questions/Q 723489
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T06:06:50+00:00 2026-05-14T06:06:50+00:00

I’m trying to do a simple task with jQuery: I have a list of

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I’m trying to do a simple task with jQuery: I have a list of words which, when hovered, should fadeIn its corresponding image. For example:

<a href="#" class="yellow">Yellow</a>
<a href="#" class="blue">Blue</a>
<a href="#" class="green">Green</a>

<img src="yellow.jpg" class="yellow">
<img src="blue.jpg" class="blue">
<img src="green.jpg" class="green">

I’m currently doing it this way for each link/image:

$('a.yellow').hover(
  function () {
    $('img.yellow').fadeIn('fast');
    },
    function () {
     $('img.yellow').fadeOut('fast');
     });

The method above works fine, but as I’m still learning, I guess there’s a better way to do that instead of repeating functions.

Can anyone give me some light here? How can I improve this code?

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  1. Editorial Team
    Editorial Team
    2026-05-14T06:06:51+00:00Added an answer on May 14, 2026 at 6:06 am
    <a href="#" class="yellow" data-type="color">Yellow</a>
    <a href="#" class="blue" data-type="color">Blue</a>
    <a href="#" class="green" data-type="color">Green</a>
    

    jQuery Code

    $('a[data-type=color]').hover(
      function () {
        $('img.'+$(this).attr("class")).fadeIn('fast');
        },
        function () {
         $('img.'+$(this).attr("class")).fadeOut('fast');
         });
    });
    

    I think you should try this one. I used data- as prefix for custom defined attributes because it is html5 compliant. You can say data-something if you want to.

    Normally you may not need to use data-color custom attribute but since I think to make it more generic, I used that attribute. You can do such code as well :

    <a href="#" class="yellow">Yellow</a>
    <a href="#" class="blue">Blue</a>
    <a href="#" class="green">Green</a>
    

    Then

    $('a').hover(
      function () {
        $('img.'+$(this).attr("class")).fadeIn('fast');
        },
        function () {
         $('img.'+$(this).attr("class")).fadeOut('fast');
         });
    });
    

    But this way, you should make sure all the links are image related links.

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