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Home/ Questions/Q 939009
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Editorial Team
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Editorial Team
Asked: May 15, 20262026-05-15T21:43:19+00:00 2026-05-15T21:43:19+00:00

I am using this code to make a certain div (#extras) fade out. $(document).ready(function()

  • 0

I am using this code to make a certain div (#extras) fade out.

$(document).ready(function() {

            for(var i = 1; i <= pageLimit; i++) {
                $('body').append('<div id="page' + i + '" name="page' + i + '" class="touch"></ div>' ); 
            }

            $('body').append('<div id="extras" class="showUp">..all content needed goes in heree...</div>');
            $(window).load(function() {
                $('div#extras').delay(100).fadeOut(4000); 
            });
});

CSS:

div#extras {
    height:100%;
}

&

body > div.showUp {
    display: block !important;
    height:100%;
}

It fades out nicely, but then pops back up once its done the fading animation.
How would I make it stay faded? In another function I will have it fade back in.

This is the defaulted css I have from the JQtouch sample I have been working on..

body > * {
background: transparent;
-webkit-backface-visibility: hidden;
display: none;
position: absolute;
left: 0;
text-align:center;
width: 100%;
-webkit-transform: translate3d(0,0,0) rotate(0) scale(1);
height: 100% !important;
}

Thanks!

Would the fact that the extras div was inserted into the page by the .append() function have anything to do with making it pop back up?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-15T21:43:19+00:00Added an answer on May 15, 2026 at 9:43 pm

    I’ve had success using:

    $(“#element”).fadeTo(“fast”, 0.33);

    The speed parameter can be set to “slow” if you want to make the fading take more time.

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