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Editorial Team
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Asked: 2026-06-17T18:34:32+00:00

I have this code: <script src=http://code.jquery.com/jquery-1.8.2.js></script> <script src=http://code.jquery.com/ui/1.9.1/jquery-ui.js></script> <script> $(function() { function animation() {

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I have this code:

<script src="http://code.jquery.com/jquery-1.8.2.js"></script>
<script src="http://code.jquery.com/ui/1.9.1/jquery-ui.js"></script>

<script>
$(function()
{
    function animation()
    {
        $('#img0').attr('src',$('#img1').attr('src')).fadeOut(4000).attr('src',$('#img2').attr('src')).fadeIn(4000).fadeOut(4000).attr('src',$('#img3').attr('src')).fadeIn(4000).fadeOut(4000) ;
        animation();
    }
});
</script>

<body>
<img  id="img0" width="613" height="260" alt="OffLease Only Lot" />
<img src="static/images/home/slides/SLIDERS-mP.jpg" id="img1" width="613" height="260" alt="OffLease Only Lot" hidden="true" />
<img src="static/images/home/slides/slider_usa.jpg" id="img2"  width="613" height="260" alt="OffLease Only Lot" hidden="true" />
<img src="static/images/home/slides/SLIDERS-ODOMETER.jpg"  id="img3"  width="613" height="260" alt="OffLease Only Lot" hidden="true" />

</body>
</html>

i want slide showing the images by changing the source of image and proceed by appearing and disappearing it.
but my code didn’t work

  • why?
  • how can i fix it?
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1 Answer

  1. Editorial Team

    If you’re trying to cycle the three images, do something like this:

    function animate1() {
    $('#img1').fadeIn(4000, function() {
        $('#img1').fadeOut(4000, animate2);
    })
}

function animate2() {
    // Do the same thing here, and do animate3 too
}

$(function() { $('#img0').fadeOut(4000, animate1) };

    I would also take a minute to go over the jQuery fadeOut and fadeIn pages. I don’t think they work the way you think they work.

    http://api.jquery.com/fadeOut/

    http://api.jquery.com/fadeIn/

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