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Home/ Questions/Q 8245221
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Editorial Team
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Editorial Team
Asked: June 7, 20262026-06-07T22:06:41+00:00 2026-06-07T22:06:41+00:00

Possible Duplicate: jQuery fade to new image I have the following code to swap

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Possible Duplicate:
jQuery fade to new image

I have the following code to swap (preloaded) images on hover:

$(document).ready(function() { 
$(".pf-item img").hover(
      function(){this.src = this.src.replace(".jpg","-h.jpg");},
      function(){this.src = this.src.replace("-h.jpg",".jpg");
 });
 var imgSwap = [];
 $(".img-swap").each(function(){
    imgUrl = this.src.replace(".jpg","-h.jpg");
    imgSwap.push(imgUrl);
});
$(imgSwap).preload();
});
$.fn.preload = function() {
this.each(function(){
    $('<img/>')[0].src = this;
});
}

It works really well, however I’d like to be able to fade between the images .jpg and -h.jpg.

Does anyone know how to do this/have a better way of approaching the issue? I have attempted lots of different fadeIn() approaches, but all seem to either not work, or work for a few hovers, then replace a blank image.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-07T22:06:43+00:00Added an answer on June 7, 2026 at 10:06 pm

    Id suggest using a mouseover/mouseout approach.

    Id have my HTML set out like this.

    <div class="imageWrap">
        <img class="shown" src="image-1.jpg" width="300" height="300" />
        <img class="hidden" src="image-2.jpg" width="300" height="300" />
    </div>
    

    Then using jQuery you can swap between them like so

    $(".imageWrap").mouseover(function() {
        $(this).children(".shown").fadeOut("fast");
        $(this).children(".hidden").fadeIn("fast");
    });
    
    $(".imageWrap").mouseout(function() {
        $(this).children(".shown").fadeIn("fast");
        $(this).children(".hidden").fadeOut("fast");
    });
    

    Dont forget your css

    .imageWrap {width:300px; height:300px;}
    .imageWrap > img {position:absolute;}
    .imageWrap > .hidden {display:none;}
    
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