Re: Re: how to detect type of image

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