PHP: exif_imagetype() not working? - php

Question

I have this extension checker:
$upload_name = "file";
$max_file_size_in_bytes = 8388608;
$extension_whitelist = array("jpg", "gif", "png", "jpeg");
/* checking extensions */
$path_info = pathinfo($_FILES[$upload_name]['name']);
$file_extension = $path_info["extension"];
$is_valid_extension = false;
foreach ($extension_whitelist as $extension) {
if (strcasecmp($file_extension, $extension) == 0) {
$is_valid_extension = true;
break;
}
}
if (!$is_valid_extension) {
echo "{";
echo "error: 'ext not allowed!'\n";
echo "}";
exit(0);
}
And then i added this:
if (exif_imagetype($_FILES[$upload_name]['name']) != IMAGETYPE_GIF
OR exif_imagetype($_FILES[$upload_name]['name']) != IMAGETYPE_JPEG
OR exif_imagetype($_FILES[$upload_name]['name']) != IMAGETYPE_PNG) {
echo "{";
echo "error: 'This is no photo..'\n";
echo "}";
exit(0);
}
As soon when I added this to my imageupload function, the function stops working. I dont get any errors, not even the one i made myself "this is no photo", what could be wrong?
Just checked with my host. They support the function exif_imagetype()

Answer 1

Here is the example with the switch usage:
switch(exif_imagetype($_FILES[$upload_name]['name'])) {
case IMAGETYPE_GIF:
case IMAGETYPE_JPEG:
case IMAGETYPE_PNG:
break;
default:
echo "{";
echo "error: 'This is no photo..'\n";
echo "}";
exit(0);
}
Your script can be used too, but you have to put AND between conditions:
if (exif_imagetype($_FILES[$upload_name]['name']) != IMAGETYPE_GIF
AND exif_imagetype($_FILES[$upload_name]['name']) != IMAGETYPE_JPEG
AND exif_imagetype($_FILES[$upload_name]['name']) != IMAGETYPE_PNG) {
echo "{";
echo "error: 'This is no photo..'\n";
echo "}";
exit(0);
}

Answer 2

There's a better, shorter way to get image types, which I'm currently using (can't be sure if it will works as it is not documented in PHP.net as image_type_to_mime_type() can also take an integer as input):
function get_image_type( $image_path_or_resource ) {
$img = getimagesize( $image_path_or_resource );
if ( !empty( $img[2] ) ) // returns an integer code
return image_type_to_mime_type( $img[2] );
return false;
}
echo get_image_type( 'somefile.gif' );
// image/gif
echo get_image_type( 'path/someotherfile.jpg' );
// image/jpeg
$image = get_image_type( 'somefile.gif' );
if ( !empty( $image ) ){
// fine, let's do some more coding
} else {
// bad image :(
}

Answer 3

if (exif_imagetype($_FILES['image']['tmp_name']) != IMAGETYPE_JPEG
AND exif_imagetype($_FILES['image']['tmp_name']) != IMAGETYPE_PNG)
{
enter code here
}
You can use like that. Hope this is will be helpful