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Answer 1

+2 A:

If you don't want to output the image directly, just don't send the header in your function. Every browser will treat the response as image file.

Also, imagedestroy($im) after return will never get executed!

If you want to just create the image and save it to a file, check the imagepng() documentation. The second parameter accepts a filename:

The path to save the file to. If not set or NULL, the raw image stream will be outputted directly.

According to your edit:

You should use imagepng with the filename parameter to create the image, and then link to it from your script.

Small example:

<?php
// ...
imagepng($im, 'foo.png');
?>
<img src="foo.png" />

The other way around would be Davide Gualanos solution by using a wrapper script to passtrough image/png headers and direct output.

Karsten 2009-11-27 13:11:37

Answer 2

+1 A:

Hi,

you can use a temporary output buffer to catch the output of the function and then having a string filled with it

Example:

ob_start();

... your image function code ...

ob_clean(); //to be sure there are no other strings in the output buffer

imagepng($im);

$string = ob_get_contents();

ob_end_clean();

return $string;

$string got all the output from imagepng();

Ass3mbler 2009-11-27 13:17:04

Hiya - I've used this, but it still returns base64 (or something) - a heap of code rather than an image, in other words. At least it's in the right place though!

hfidgen 2009-11-27 13:22:54

have you tried to $string = base64_decode($string) to see if you've got the original image?

Ass3mbler 2009-11-27 13:27:44

Answer 3

+1 A:

You have to put the script that output the image in a page, and call that page in a html tag to display it.

Example:
image.php

<php
$image = $_GET['image'];
$text = $_GET['text'];
makeimage($image, $text);
?>

page.html:

<html>
<body>
<img src="image.php?image=foo.jpg&text=hello" />
</body>
</html>

Davide Gualano 2009-11-27 13:17:59

Ahh perfect - that makes sense!

hfidgen 2009-11-27 13:38:55

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