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Editorial Team
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Asked: 2026-05-27T06:26:36+00:00

I am not very familiar with java-script/ jquery. i am creating a sliding social

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I am not very familiar with java-script/ jquery. i am creating a sliding social widget for my blog.
though i am able to place it on a exact location but it is not moving as i am scrolling the page,after some goggling and checking through firebug i am seeing following error in the firebug console.

$(".floater").floatingFixed is not a function

here is the code which i have written so far for this slider option

    <script type="text/javascript">
              $(function() {
                $(".floater").floatingFixed({ padding: 5 });
              });
            </script>
            <style type="text/css">
              #widgetcontainer { padding: 50px 220px; text-align: justify; }
              .floater { background: none repeat scroll 0 0 #C9DBEC; position: absolute; top: 215px; left: 150px; width: 58px; padding:4px 4px 0;text-align:center }
              .post_share{-moz-border-radious:4px 4px 4px 4px;background:none repeat scroll 0 0 #C9DBEC;padding:4px 4px 0;z-index:11500}

            </style>

<div id="widgetcontainer">
    <div class="floater">
button options

</div>

i am unable to find out the cause of this error in firebug.Any help in this regard will be really helpful

thanks in advance

Update
here are the js file being included in the header section

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.2/jquery.min.js" type="text/javascript"></script>
    <script src="<?php echo get_option('home'); ?>/scripts/jquery.floatingFixed.js" type="text/javascript"></script>
    <script src ="<?php echo get_option("siteurl"); ?>/js/jquery.js" type="text/javascript"></script>
    <script src ="<?php echo get_option("siteurl"); ?>/js/jquery.simplemodal.js" type="text/javascript"></script>
    <script src ="<?php echo get_option("siteurl"); ?>/js/contact.js" type="text/javascript"></script>
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1 Answer

  1. Editorial Team

    The problem is your referencing the jQuery core multiple times – the second time after your plugin .. this causes the $ to be recreated

    <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.2/jquery.min.js" type="text/javascript"></script>
<script src="<?php echo get_option('home'); ?>/scripts/jquery.floatingFixed.js" type="text/javascript"></script>
<script src ="<?php echo get_option("siteurl"); ?>/js/jquery.simplemodal.js" type="text/javascript"></script>
<script src ="<?php echo get_option("siteurl"); ?>/js/contact.js" type="text/javascript"></script>

    Removing the duplicate jQuery should do the trick.

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