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Home/ Questions/Q 8878863
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Editorial Team
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Editorial Team
Asked: June 14, 20262026-06-14T19:48:57+00:00 2026-06-14T19:48:57+00:00

I want to add .delay() to this, so each item will animate one after

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I want to add .delay() to this, so each item will animate one after the other. The problem is that if I add delay() to the element the fadeIn stops working.

Working code (but without delay…)

time = 500;

for (var i=1;i<=5;i++){

    delay2 = (i * time);
    $('<tr><td><h3>Hello</h3></td><td>'+i+'</td</tr>').hide().appendTo('#table').fadeIn("slow").css('display', 'table-row');

    // do more stuff here

};

jsfiddle example

FadeIn not working (as it has delay…)

time = 500;

for (var i=1;i<=5;i++){

    delay2 = (i * time);
    $('<tr><td><h3>Hello</h3></td><td>'+i+'</td</tr>').hide().appendTo('#table').delay(delay2).fadeIn("slow").css('display', 'table-row');

    // do more stuff here

};

jsfiddle example

Does anyone know what is the problem? In the second example it should animate the items one after te other, but that does not happen, they’re not even animated.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-14T19:48:58+00:00Added an answer on June 14, 2026 at 7:48 pm

    Try this:

    $('<tr><td><h3>Hello</h3></td><td>'+i+'</td</tr>')
    .appendTo('#table')
    .hide()
    .delay(delay2)
    .show('slow');
    

    The problem here is that the css change occurs instantly, whereas you want it to occur after the fade in is complete. You do not need fadeIn at all here, especially since show will remember the display attribute value and restore it automatically.

    Here is a fiddle: http://jsfiddle.net/u5dEp/7/

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