Date: 03/24/07 (PHP Community)    Keywords: database

I've been testing out image uploading and resizing on the fly tonight..
While I have the page able to pull the image name from the database. It seems like the script I've found can't display the picture, as a variable, the 'imagejpeg' shows the image fine but how do I end the script so that I can have the sized image name in a variable.




        $gvsql = "SELECT * FROM image_gallery WHERE gid = \"$gid\" 
                             ";
        $gvresult = @mysql_query($gvsql,$conn) or die("Couldn't execute query.");
 
        while ($gvrow = mysql_fetch_array($gvresult)) {
         
                         $img = $gvrow['image_name_resize'];
                                 
                                        }
                                       
        $filename = '/home/benjamin/www/dev/upload/';

        $width = 300;
        $height = 300;

        header('Content-type: image/jpeg');

          $filefile="$filename$img";
        list($width_orig, $height_orig) = getimagesize($filefile);

        $ratio_orig = $width_orig/$height_orig;

        if ($width/$height > $ratio_orig) {
           $width = $height*$ratio_orig;
        } else {
           $height = $width/$ratio_orig;
        }
        $image_p = imagecreatetruecolor($width, $height);
        $image = imagecreatefromjpeg($filefile);
        imagecopyresampled($image_p, $image, 0, 0, 0, 0, $width, $height, $width_orig, $height_orig);
        imagejpeg($image_p, null, 100);
        imagedestroy($image_p);
?>

Source: http://community.livejournal.com/php/556781.html