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Home/ Questions/Q 8618243
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Editorial Team
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Editorial Team
Asked: June 12, 20262026-06-12T06:03:39+00:00 2026-06-12T06:03:39+00:00

With the html: <a class=jslider-pointer href=# style=left: 50%;></a> I used jquery api css() to

  • 0

With the html:

<a class="jslider-pointer" href="#" style="left: 50%;"></a>

I used jquery api css() to get the value of left property:

$('.jslider-pointer').first().css('left')

It worked well in chrome, which returned 50%. But it didn’t work in Firefox. It returned 150px in Firefox.

Does anyone know why it happened and how could I get the right value in Firefox(a method working across different browsers)?

Thanks

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  1. Editorial Team
    Editorial Team
    2026-06-12T06:03:39+00:00Added an answer on June 12, 2026 at 6:03 am

    According to this other post Retrieving percentage CSS values (in firefox) this is a know bug. Based on the fiddle in that post’s answer I’ve made a jQuery plugin which should get rid of your problem:

    (function($) {
        $.fn.cssFix = function(property) {
            var el = this[0], value;
    
            if (el.currentStyle)
                value = el.currentStyle[property];
            else if (window.getComputedStyle)
                value = document.defaultView.getComputedStyle(el,null).getPropertyValue(property);
    
            return value;
        }
    })(jQuery);
    

    Usage:

    $('.jslider-pointer').first().cssFix('left');
    

    Here’s a fiddle that uses this plugin and works on both Chrome and Firefox, and returns the value as it was defined in the css rule: either % or px.

    EDIT: Tested in Internet Explorer and it works there as well

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