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Home/ Questions/Q 6546741
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Editorial Team
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Editorial Team
Asked: May 25, 20262026-05-25T11:44:43+00:00 2026-05-25T11:44:43+00:00

This will animate the second image after the first animation is complete: <script type=text/javascript>

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This will animate the second image after the first animation is complete:

<script type="text/javascript">
        $(document).ready(function () {

            $('#image1').show("slide", { direction: 'left' }, 1000,
             function() { $('#image2').show("slide", { direction: 'left' }, 1000)});

    </script>

If I have four images I tried below – and only the first two images animate.

<script type="text/javascript">
        $(document).ready(function () {

            $('#image1').show("slide", { direction: 'left' }, 1000,
             function() { $('#image2').show("slide", { direction: 'left' }, 1000)},
             function () { $('#image3').show("slide", { direction: 'left' }, 1000)},
             function () { $('#image4').show("slide", { direction: 'left' }, 1000)});

    </script>

So what would be an effective way to animate all the images (4) in sequence?

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  1. Editorial Team
    Editorial Team
    2026-05-25T11:44:43+00:00Added an answer on May 25, 2026 at 11:44 am
    $('#image1').show("slide", { direction: 'left' }, 1000, function() {
        $('#image2').show("slide", { direction: 'left' }, 1000, function () {
            $('#image3').show("slide", { direction: 'left' }, 1000, function () { 
                $('#image4').show("slide", { direction: 'left' }, 1000);
            });
        });
    });
    

    you could also do:

    slide(1);
    function slide(id){
        if(id<5){
            $('#image'+id).show("slide", { direction: 'left' }, 1000, function(){ slide(id+1));
        }
    }
    
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