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Home/ Questions/Q 6663579
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Editorial Team
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Editorial Team
Asked: May 26, 20262026-05-26T02:30:53+00:00 2026-05-26T02:30:53+00:00

im getting a little issue here. Im trying to load an image, but seems

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im getting a little issue here. Im trying to load an image, but seems it isnt correct. Im getting the data of image, but i just want to load it instead of.

Whats wrong?

var url = $(this).find("img").attr("ref");

$("#produto .main-photo .photo img").load(url, function() {

                    $("#produto .main-photo .loader").fadeOut();
                    $("#produto .main-photo .photo").fadeIn();

});
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  1. Editorial Team
    Editorial Team
    2026-05-26T02:30:53+00:00Added an answer on May 26, 2026 at 2:30 am

    You should not use load for this. Simply set the src of the image:

    $("#produto .main-photo .photo img").attr("src",url);
    

    load is used to make an AJAX call.

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