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Home/ Questions/Q 6847117
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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T00:42:21+00:00 2026-05-27T00:42:21+00:00

$(‘img.clientImage’).live(‘hover’, function () { if ($(‘div#ClientImageHover’).length > 0) { $(‘div#ClientImageHover’).remove(); } else { $(‘<div

  • 0
$('img.clientImage').live('hover', function () {
    if ($('div#ClientImageHover').length > 0) {
        $('div#ClientImageHover').remove();
    } else {
        $('<div id="ClientImageHover">Change Image</div>').insertAfter($(this));
        $('div#ClientImageHover').css({ 'top': $(this).position().top });
    }
})

Now what happens if I hover over #ClientImageHover. You guessed it, it will start flickering quickly on and off. Because there’s a mouseout event on .clientImage.
I need to create the element here and append it after the img then position it on top of it. This is working correctly, but I am having issues when hovering over #ClientImageHover. How can I keep showing this div normally when the mouse is over it and keep everything as it currently is?

Thanks.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-27T00:42:22+00:00Added an answer on May 27, 2026 at 12:42 am

    To expand on my comment, do something like this jsFiddle

    HTML:

    <div class="container">
        <img class="clientImage"></div>
    </div>
    

    JS

    $('.container').live('hover', function () {
        if ($('div#ClientImageHover').length > 0) {
            $('div#ClientImageHover').remove();
        } else {
            $('<div id="ClientImageHover">Change Image</div>').appendTo($(this));
            $('div#ClientImageHover').css({ 'top': $(this).position().top });
        }
    }); 
    
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