Question

Im trying to determine the mime type of an image:

$image = $_FILES['image'];  //code shortened
function determineImage($imageResource){

    $errors = array();

    $types = array('gif' => IMAGETYPE_GIF,
                   'jpeg' => IMAGETYPE_JPEG,
                   'png' => IMAGETYPE_PNG,
                   'bmp' => IMAGETYPE_BMP);


    if ( !in_array(exif_imagetype($imageResource['tmp_name']), $types )) {
        $errors[] =  'Cannot determine mime type';
    }

    if ($imageResource['type'] !== 'image/gif'   ||
        $imageResource['type'] !== 'image/jpeg'  || 
        $imageResource['type'] !== 'image/pjpeg' ||
        $imageResource['type'] !== 'image/png'){
            $errors[] = 'Again cannot determine type';
    }

    return $errors;

}

I use

   var_dump(determineImage($image)); 

this keep returning array(1) { [0]=> string(27) “Again cannot determine type” }

However this:

echo $image['type'];

just returns:

image/png

I’ve also got error_reporting(E_ALL) turned on. Can anyone make out what the problem is, have I made a silly mistake?

Answer 1

the code is a total random (the check doesn’t make sense at all)

by the way

if ($imageResource['type'] !== 'image/gif'   AND
    $imageResource['type'] !== 'image/jpeg'  AND
    $imageResource['type'] !== 'image/pjpeg' AND
    $imageResource['type'] !== 'image/jpg'  ...

AND not OR