Verifying MIME type

apaxson

I am building an image viewer where PHP returns a list of the images in a directory, and passes that on to Flash, where the images are loaded dynamically.

My question is, is there a way to verify that the current file that PHP is reading is of type image/jpeg? In other words, if I have a text file, a gif file, and an HTML file with all my JPEG's in that directory, I want PHP to skip everything that is not image/jpeg?

Does PHP support such a function? I'm thinking my best bet is to use getimagesize() since that will return the correct MIME-type. And if it's not, it will return NULL and generate an error. Can I supress this with '@'? I've never used this function before.

Can anyone give some ideas on the best way to go about this?

lnfreish

$dir = "c:/my pictures";
$handle = opendir($dir);

while (($file= readdir($handle)) != false) {
$extension = strtolower(strrchr($file));
if (($extension == ".jpg") or ($extension == ".jpeg") {
showimage();
}
}

this way will verify the JPEG file by extension.

suntra

I found this...

<? function LoadJpeg ($imgname) {
    $im = @imagecreatefromjpeg ($imgname); /* Attempt to open */
    if (!$im) { /* See if it failed */
        $im  = imagecreate (150, 30); /* Create a blank image */
        $bgc = imagecolorallocate ($im, 255, 255, 255);
        $tc  = imagecolorallocate ($im, 0, 0, 0);
        imagefilledrectangle ($im, 0, 0, 150, 30, $bgc);
        /* Output an errmsg */
        imagestring ($im, 1, 5, 5, "Error loading $imgname", $tc);
    }
    return $im;
} ?>

apaxson

Thanks lnfreish,

but I couldn't get your code to work. however, you did get me on the right lines of thinking. Here's my code:

<? //retrieve path to current script
$path = $_SERVER["PATH_TRANSLATED"];
//retreive directory that the current script is in
$currentdir = dirname ($path);
//open directory (create handle)
$dir = opendir($currentdir);
//read through the files of the current directory
while ($file = readdir($dir)) {
	$extension = strtolower(strstr($file,".")); 
	if ($extension == ".jpg" || $extension == ".jpeg") { 
	echo "$file<br>";
	}
}

Sometimes I just think of things too difficult. You're right, it's much easier to base it on extension! Thanks!!

Suntra,

Thanks too! But, unfortunately, PHP is doing nothing more, than returning the filenames of the images. Flash is the one that is opening and manipulating them. But, I can see a use for this later! Thanks!