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Home/ Questions/Q 6070815
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Editorial Team
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Editorial Team
Asked: May 23, 20262026-05-23T09:58:34+00:00 2026-05-23T09:58:34+00:00

I’m using jShowOff in a WordPress theme inside home.php within the <body> . Inside

  • 0

I’m using jShowOff in a WordPress theme inside home.php within the <body>.

Inside the <head> in header.php I have both jQuery and jShowOff scripts properly included:

<script type='text/javascript' src='https://ajax.googleapis.com/ajax/libs/jquery/1.6.1/jquery.min.js?ver=1.6.1'></script>
<script type="text/javascript" src="../js/jquery.jshowoff.min.js"></script>

but the jShowOff functionality only works if I also include jQuery inside home.php (within the same php file as where I’m putting jShowOff HTML code).

Can anyone think of why this might be or how I can fix it? I shouldn’t need to load jQuery twice, right?

Within the <body> in home.php the code looks like this:

<script type='text/javascript' src='https://ajax.googleapis.com/ajax/libs/jquery/1.6.1/jquery.min.js?ver=1.6.1'></script>

            <div id="basicFeatures">
                <div title="Shore"><a href="http://google.com"><img src="http://farm5.static.flickr.com/4017/4303103738_a4745a3e6d_o.jpg" width=660 height=371 alt="Shore"></a></div>
                <div title="Flower"><a href="http://imgur.com"><img src="http://farm5.static.flickr.com/4067/4302354517_d72d321f17_o.jpg" width=660 height=371 alt="Flower"></a></div>
                <div title="Fern"><a href="http://gmail.com"><img src="http://farm3.static.flickr.com/2739/4303103822_a3b23ff7f5_o.jpg" width=660 height=371 alt="Fern"></a></div>
            </div>
            <script type="text/javascript">     
                $(document).ready(function(){ $('#basicFeatures').jshowoff({ 
                    links: false,
                    controls: false,
                    effect: 'fade',
                    cssClass: 'basicFeatures',
                    hoverPause: false 
                }); });
            </script>

Update: Problem solved by switching $ to jQuery:

        <script>        
            jQuery(document).ready(function(){ jQuery('#basicFeatures').jshowoff({ 
                links: false,
                controls: false,
                effect: 'fade',
                cssClass: 'basicFeatures',
                hoverPause: false 
            }); });
        </script>
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-23T09:58:35+00:00Added an answer on May 23, 2026 at 9:58 am

    Remove the script include tag in your page’s body. Once the page has loaded, open javascript console of your browser and check whether it responds to jQuery and $ both should respond the same object. If they doesn’t it means your $ used by some other library. Replace $ with jQuery.

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