no no no no no that's not how it works. You cant just call a function like that and it will display an image - think about it in terms of the PHP creating html - you need an <img src=...> tag to get an image, and that's not what that function creates.
What you need to do is create a makeimage.php file (i've posted the one i use for jpegs just for you to look at) and then call it from your main php file like so:
echo "<img src=\"makeimage.php?directory=1&file=this.jpg\">
here's my makeimage.php file:
<?
// if (!$max_width)
$max_width = 100;
// if (!$max_height)
$max_height = 100;
$file = $directory."/".$file;
$size = GetImageSize($file);
$width = $size[0];
$height = $size[1];
$x_ratio = $max_width / $width;
$y_ratio = $max_height / $height;
if ( ($width <= $max_width) && ($height <= $max_height) ) {
$tn_width = $width;
$tn_height = $height;
}
else if (($x_ratio * $height) < $max_height) {
$tn_height = ceil($x_ratio * $height);
$tn_width = $max_width;
}
else {
$tn_width = ceil($y_ratio * $width);
$tn_height = $max_height;
}
$src = ImageCreateFromJpeg($file);
$dst = ImageCreate($tn_width,$tn_height);
ImageCopyResized($dst, $src, 0, 0, 0, 0,
$tn_width,$tn_height,$width,$height);
header("Content-type: image/jpeg");
ImageJpeg ($dst, '', 90);
ImageDestroy($src);
ImageDestroy($dst);
?>
given as your code is generating an image, putting things like echo into it dont do anything or will confuse it - they'll have to go in your main script.