My problem is that instead of treating the code as html it is simply printing it out... is it possible to have HTML inside a generate PHP Image?

<?php
include "includes/config.php";
Header ("Content-type: image/png");
$img_handle = imageCreateFromPNG("images/blank.png");
$color = ImageColorAllocate ($img_handle, 10, 100, 100);

$image = '<img src="images/emo.gif" />';

ImageString ($img_handle, 3, 10, 9, "$image", $color);
ImagePng ($img_handle);
ImageDestroy ($img_handle);
?>

Thanks :)

If you look at this:
Header ("Content-type: image/png");
You can see that it can ONLY output an image. Not HTML. Not data. Nothing else. If you could tell me a little more about what you need it for, I may more easily understand your intentions here.

Isnt this exactly what GD does, you just need to use the correct PHP code to do so.

if you want to retrieve a image in php generated html code you would need another php file to get the image

image.php:-
******************************
<?php
include "includes/config.php";
header("Content-type: image/png");
$img_handle = imageCreateFromPNG("images/blank.png");
$color = ImageColorAllocate ($img_handle, 10, 100, 100);

$image = '<img src="images/emo.gif" />';

ImageString ($img_handle, 3, 10, 9, "$image", $color);
ImagePng ($img_handle);
ImageDestroy ($img_handle);
?>
**************************************

now make a html or php html any thing and where you have the img tag

<img src=image.php>

Try this:


<?php
include "includes/config.php";
header ("Content-type: image/png");
$img_handle = imageCreateFromPNG("images/blank.png");
$color = ImageColorAllocate ($img_handle, 10, 100, 100);

$image = 'images/emo.gif';

ImageString ($img_handle, 3, 10, 9, $image, $color);
ImagePng ($img_handle);
ImageDestroy ($img_handle);
?>


Peace,