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
– zaf, Commented Jan 31, 2011 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
– Your Common Sense, Commented Jan 31, 2011 at 16:07
My poor boy, to get a detailed answer one have to ask a detailed question! And sensible one. — Your Common Sense
– Your Common Sense, Commented Jan 31, 2011 at 16:08
If you have nothing constructive or helpful to say, you are usually better off keeping your mouth shut. — Arc
– Arc, Commented Jan 31, 2011 at 16:14

Pekka · Accepted Answer · 2011-01-31 16:07:08Z

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, 2011 at 16:07

Pekka

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1 Comment

Arc Over a year ago

Which will only work if the php.ini setting allow_url_fopen is enabled. php.net/manual/en/…

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