You could also use the function getimagesize this will tell what kind of
image it is, and if it is not an image it will return false.
GD LIB is NOT needed for getimagesize.
From: Daniel Grace [mailto:[EMAIL PROTECTED]]
Sent: Monday, December 17, 2001 6:52 AM
To: [EMAIL PROTECTED]
Subject: Re: problem finding out original filename while using php to
"Neil M" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> I am using php to upload image files.
> heres my code ..
> if (is_uploaded_file($photo_file))
> clearstatcache ();
> $size = filesize ( $photo_file ) ;
> if ( $size > 1024000 )
> print ("<BR> ERROR - File is too big ! File uploads should be below 1024
> ( 1mb ) <BR>"); exit ;
> move_uploaded_file($photo_file, "$upload_path$check_nick->nickname");
> echo "<BR> ERROR : File upload was not successfull , please try again
> the problem is that when i look for the original filename ( e.g.
> myphoto.gif ) , $photo_file contains a random file name like 383hr93php
> As i mentioned , its image files i am uploading , how do i know what type
> image the file was ? like .jpg , .png etc.
> Any help appreciatted , i am really stuck on this ;0)
> Neil M
$photo_file will map to the name of the temporary file on the server's HD --
it is NOT the name of the original file. You'll want to use $photo_file_name
to determine the original filename, or better yet, $photo_file_type to
determine the content type. (a gif is image/gif, for instance).
echo make_witty_sig($foo) . "\n";
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