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Home/ Questions/Q 8446209
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Editorial Team
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Editorial Team
Asked: June 10, 20262026-06-10T09:52:07+00:00 2026-06-10T09:52:07+00:00

I have a PHP script that generates an image with PHP GD. After it

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I have a PHP script that generates an image with PHP GD. After it generates the image, it saves it, and send this output when called by Ajax:

imagejpeg($img_data, 'filename.jpg');
echo '<img src="/filename.jpg.jpg">';

And after that, the image is shown on the page, and everything is fine. But, I don’t want to create an image every time. Is there some way that I return by Ajax only $raw_data string and show the image? I tried like this:

echo $img_data;

But no luck, only thing that is returned is a few ?.

Here is my jQuery Ajax code:

$.ajax({
  type: 'POST',
  data: {
    action: 'update_image',
    //some instructions for creating the image
  },
  url: 'script.php',
  success: function(msg) {
    $('#somediv').append(msg);
  }
});
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-10T09:52:09+00:00Added an answer on June 10, 2026 at 9:52 am

    base64 encode the image and return that, then you can do an <img src="data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAADIA......." />

    On the PHP side

    $image = base64_encode($imageGDRender);
    echo json_encode(array('image'=>$image));
    

    That will return your json back to your jquery

    then on the ajax side

    $.ajax({
        ...
        success: function(data) {
            var base64Image = data.image;
            ...now put it in your image
            $('#image').attr('src','data:image...'+base64Image);
        })....
    
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