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Home/ Questions/Q 6348869
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Editorial Team
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Editorial Team
Asked: May 24, 20262026-05-24T21:30:21+00:00 2026-05-24T21:30:21+00:00

I’m working on http://www.aceaofmidland.com/ and I’m trying to use jQuery and jQuery.cycle to cycle

  • 0

I’m working on http://www.aceaofmidland.com/ and I’m trying to use jQuery and jQuery.cycle to cycle through a wordpress custom post type that’s marked up like so:

<div id="toolbar">
    <div id="home">
        <div id="arrows">
            <div class="prev"></div>
            <div class="next"></div>
        </div>
        <div id="descriptions">
            <div><strong>Second Home Alert Test</strong><br><p>Testing Testing Testing</p></div>
            <div><strong>Test Home Alert Title</strong><br><p>Test Content for homepage alert</p></div>
    </div>
    </div>
</div>

I’m using the following jQuery to try and activate the cycling. I’m positive I have the scripts correctly loaded, and in order too.

<script type="text/javascript">
    jQuery.noConflict();
    jQuery(function(){  
        jQuery('#descriptions').cycle({
            next: '.next',
            prev: '.prev',
            fx: 'fade',
            speed: 'fast',
            timeout: 5000,
            pause: 1
        });
    });
</script>

If anyone can find out how to get this to work, I’ll be forever grateful. Thank you!

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-24T21:30:22+00:00Added an answer on May 24, 2026 at 9:30 pm

    In firebug I get the following errors on pageload:

    jQuery.cookie is not a function
    http://aceaofmidland.com/
    Line 126
    

    and

    $j.cookie is not a function
    http://aceaofmidland.com/wp-content/themes/trademark/files/js/theme.js?ver=3.2.1
    Line 570
    

    You may want to change this line:

    if(undefined != jQuery.cookie('links-variant')) {
    

    to

    if('undefined' != typeof(jQuery.cookie('links-variant'))) {
    

    https://developer.mozilla.org/en/JavaScript/Reference/Operators/Special/typeof

    And that jQuery code should probably be wrapped in a $(document).ready(function () {}); wrapper.

    I would always try to resolve all errors that occur before the code I am troubleshooting before altering the problematic code itself.

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