Stupid question time! I can't find my answer anywhere, sorry about this...
The nature of the return value of exif_imagetype() is baffling me. The
php manual uses one simple example like this:
<code>
<?php
if (exif_imagetype('image.gif') != IMAGETYPE_GIF) {
echo 'The picture is not a gif';
}
?>
</code>
okay, cool. the return value is the constant IMAGETYPE_GIF. of course
if you do something like this:
<code>
$foo = exif_imagetype('image.jpg');
echo $foo;
</code>
you'll echo out 2.
Great, seems really useful to me. What am I missing? Why is the value
of the IMAGETYPE_XXX constants an integer, for one thing, and not a
string like 'gif' or 'jpg'? And how do I access some sort of usable
string telling me the type of image without having to do something
retarded like putting a bunch of strings in a numerically indexed array
(which will be completely confusing later when I try to figure it out).
And thirdly, what the !?#@ am I missing here? I feel like an idiot, but
it makes no sense to me that an 'imagetype' function would return
something that evaluates to an integer.
Please help, my head hurts.
Thanks
ec
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