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Home/ Questions/Q 4612924
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Editorial Team
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Editorial Team
Asked: May 22, 20262026-05-22T01:28:20+00:00 2026-05-22T01:28:20+00:00

this is my code on jsFiddle, when I try doing $(‘#featuredStores’).fadeOut(1000) it works, but

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this is my code on jsFiddle, when I try doing

$('#featuredStores').fadeOut(1000)

it works, but .fadeIn() doesn’t work, Similarly when I use .slideUp() it works, but doesn’t work when I use .slideDown()

what’s the issue here?

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  1. Editorial Team
    Editorial Team
    2026-05-22T01:28:20+00:00Added an answer on May 22, 2026 at 1:28 am

    Basically slideDown() shows the image. Considering your image is already shown when the DOM is rendered, there is nothing to slide down to. If you set the style as display:none then use slideDown(), it’ll work.

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