Ajax and returning image created by PHP GD

Question

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);
  }
});

romo · Accepted Answer · 2012-08-27T15:46:11.610

5

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);
    })....

edited Aug 27 '12 at 15:46

answered Aug 27 '12 at 14:57

romo

1,990
11
10

I am unsure how to implement this? As I can google about this, i need some "json encoding" to achieve this or something like that? – SomeoneS Aug 27 '12 at 15:32
Tnx, but still instead of image data, i get "undefined". But this helped me to search in right place. I put solution in answer below. – SomeoneS Aug 27 '12 at 16:20
Note that you'll lose caching – Michael Laffargue Aug 28 '12 at 13:26

score 2 · Answer 2 · answered Aug 27 '12 at 16:19

Here is solution:

PHP code:

ob_start();
imagejpeg($im);
$outputBuffer = ob_get_clean();
$base64 = base64_encode($outputBuffer);
echo '<img src="data:image/jpeg;base64,'.$base64.'" />';

On user side (in JQuery):

success: function(data) {
        $('#somediv').append(data);
}

score 0 · Answer 3 · answered Aug 27 '12 at 15:01

0

You should modify the header before doing the echo $img_data;

header("Content-Type: image/jpeg");
header("Content-Length: " .sizeof($img_data));

answered Aug 27 '12 at 15:01

Michael Laffargue

10,116
6
42
76

Result is same when i do this, nothing changes. – SomeoneS Aug 27 '12 at 15:30

Ajax and returning image created by PHP GD

3 Answers3

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