PHP imagecreatefrompng(file) fails when file is a dynamic image file?

Question

Is it possible to use imagecreatefrompng() with a php file that returns a dynamic png image?

eg.

<?php
 $IM = imagecreatefrompng('image.php?var=1');
?>

where image.php looks something like:

<?php
 // code to generate image
 header("content-type: image/png");
 imagepng ( $OUTPUT );
?>

At the moment I'm getting a "failed to open stream" error - can this be done? If not, are there any quick & easy workarounds? (that don't involve using image.php to save a .png file for the script to detect instead.)

Thanks,

Chris

@Col. Shrapnel Oh dude, please help the guy out by giving a detailed answer! — zaf, Jan 31 '11 at 16:06
why so complex structure? ever tried to create an image ONCE, not thousand times, passing it from one script to another? — Your Common Sense, Jan 31 '11 at 16:07
My poor boy, to get a detailed answer one have to ask a **detailed question!** And sensible one. — Your Common Sense, Jan 31 '11 at 16:08
If you have nothing constructive or helpful to say, you are usually better off keeping your mouth shut. — Arc, Jan 31 '11 at 16:14
*sigh* I *knew* my answer was going to get a downvote from @Col. Shrapnel. After all, answering questions he doesn't approve of is *wrong* — Pekka, Jan 31 '11 at 16:16
@Pekka you know that requesting urls from your own site is just stupid. And you know that making such comment will never help - these people *always* knows better, what to to. — Your Common Sense, Jan 31 '11 at 16:19

score 4 · Accepted Answer · answered Jan 31 '11 at 16:07

4

Query parameters ?var=1 work only when requesting a resource through http, not through the file system. To do this, you would have to specify a full URL:

<?php
 $IM = imagecreatefrompng('http://localhost/image.php?var=1');
?>

(If your PHP is configured to allow this)

However, the usually much more desirable way would be to include image.php directly and pass var to it using a normal variable. That saves you a http request that, even if made locally, spawns a new PHP process.

answered Jan 31 '11 at 16:07

Pekka

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Which will only work if the `php.ini` setting `allow_url_fopen` is enabled. http://php.net/manual/en/filesystem.configuration.php#ini.allow-url-fopen – Arc Jan 31 '11 at 16:09

PHP imagecreatefrompng(file) fails when file is a dynamic image file?

1 Answers1