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Home/ Questions/Q 6911315
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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T08:56:13+00:00 2026-05-27T08:56:13+00:00

Simple concept, but I’m having trouble coming up with an elegant solution. I’ve got

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Simple concept, but I’m having trouble coming up with an elegant solution.

I’ve got a series of images in a gallery and the amount inside will change. I’ve written a short script that runs through a slideshow, fading the visible image out with the next underneath it. Conceptually it’s perfect, but above 2 images it gets wonky because the “next” images are all on the same z-index.

Is there a simple script that can count the photos and add a progressive z-index to them? For example: there are 10 images in the gallery and it assigns them z-indexes -1 through -10.

Thanks for your help!

jsFiddle: http://jsfiddle.net/danielredwood/mmPZN/

HTML:

<img class="slide first" src="img/diamond-test.jpg" />
<img class="slide" src="img/diamond-test2.jpg" />
<img class="slide" src="img/diamond-test3.jpg" />

CSS:

.slide {
  top:100px;
  position:absolute;
  z-index:-1;
}
.first {
  display:block;
  position:static;
}

JavaScript:

$('.slide').click(function(){
    var pic = $(this);
    pic.fadeOut(400, function(){
        pic.next().addClass('first');
        pic.insertAfter('.slide:last').removeClass('first').show();
    });    
});
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-27T08:56:14+00:00Added an answer on May 27, 2026 at 8:56 am

    Figured out a solid fix, didn’t need to z-index all the images sequentially.

    See here: http://jsfiddle.net/danielredwood/mmPZN/1/

    JavaScript:

    $('.slide:first').addClass('first').next().addClass('second');
    $('.slide').click(function(){
        var pic = $(this);
        pic.fadeOut(400, function(){
        pic.next().removeClass('second').addClass('first').next().addClass('second');
            pic.insertAfter('.slide:last').removeClass('first').show();
        });    
    });
    
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