On Apr 29, 2007, at 8:03 AM, tedd wrote:
At 4:46 PM -0500 4/28/07, Edward Vermillion wrote:
On Apr 28, 2007, at 12:21 PM, tedd wrote:
At 9:22 AM -0500 4/28/07, Edward Vermillion wrote:
It should, but instead try this:
$image_size = getimagesize($filename);
echo $image_size['mime'];
$image_size['mime'] ? Where did that come from?
I duno, maybe the manual.
http://us2.php.net/getimagesize <-- 5th or 6th example down.
Ahhh... that's for sending a mime type to the browser.
Ahhh no, it's just a way to get contents of the file.
If you really want to study this, try opening every different image
file (gif, jpg, png, etc.) you have on your desktop and examine
each of the header files via a HEX editor. You will find that every
file has an id of some type in it's header.
The php functions that provide data about files, do just that. They
inspect the header of the file and report what they have found. How
you use them, is your business.
Well, from the example it looks like that's returning a string that
can go straight into the header() function...
<?php
$size = getimagesize($filename);
$fp=fopen($filename, "rb");
if ($size && $fp) {
header("Content-type: {$size['mime']}");
fpassthru($fp);
exit;
} else {
// error
}
?>
whereas index 2 in the array returns an integer that corresponds to
the IMAGETYPE_* constants. I would assume they both get the
information from the same place, but just return it in different
formats based on the intended usage. Just like index 3 will get you a
string of 'width=? height=?' to add to an image tag, while index 0
and 1 returns an integer of width and height respectively.
I can get the width and height from index 3, but that's not what it
was designed for.
Ed
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