PHP render html code to image

Question

render.php

$form = '<b>Hi</b><br /><img src="test.jpg" /><br />New File';

echo $form;
header("Content-Type: image/png");
imagejpeg($form);

But this code not work

I want this page render to img file :

<img src="render.php" />

Remove last 2 lines, from your render.php. See if that helps. — Gaurav Dave, Jan 17 '15 at 12:10

score 0 · Answer 1 · edited May 23 '17 at 12:27

0

you are trying to output both HTML and image-data - that doesnt make any sense. Instead you should put an HTML-img-tag into your website, the src-attribute being render.php which sets the header via header() and directly outputs image data - without any additional HTML.

https://stackoverflow.com/a/1851856/351861

edited May 23 '17 at 12:27

Community

1
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answered Jan 17 '15 at 12:14

specializt

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Thx , can you help more ? – Mohammad Jan 17 '15 at 12:19
there is nothing more to help you with, you now have a full solution to your problem. If you have more problems you should create another question. – specializt Jan 17 '15 at 12:20
Thx , But this answer have $file : $file = '../image.jpg'; . my code have not image.jpg – Mohammad Jan 17 '15 at 12:39

score 0 · Answer 2 · answered Jan 17 '15 at 12:39

You should use extra php file for images, something like:

<img src="render.php?id=15" />

Because if your php script returns only one picture it is used wrong. so you pass it what image to show. And render.php might be something like

$pic_id = (int)$_GET['id'];// cast to int because only numbers are allowed here
$pic_path = 'assets/img/';
$f = $pic_path.$pic_id.".jpg";// it's just an example so it's simplyfied
header("Content-Type: image/png");
imagejpeg($form);

The point here is that you can check user's permissions to see the picture very flexible and log some data about downloader.

PHP render html code to image

2 Answers2