[PHP-DB] One more GD ? -- filename

[Craig Hoffman]

Thanks everyone for answering my other post.  However, I'm still having 
trouble passing the image name from the GD script to the image tag.  
I'm including the GD script below. Any help would be great. Thanks - 
Craig

//GD script
<?php
	include ("db.php");
	$query = "SELECT route_photo FROM routes WHERE route_photo IS NOT NULL 
ORDER BY RAND() LIMIT 1";
	$result = mysql_query($query, $db) or die(mysql_error());

while($row = mysql_fetch_array($result)) {
	
	//Photo varibale
		$route_photo = $row["route_photo"];
		
	//Start GD
	//New Image resize
		$new_width = 125;
		$new_height = 225;
		
	//Content Type
		header("Content-type: image/jpeg");
		
	//Start new image resize	
		list($width_orig, $height_orig) = 
getImageSize("../images/climbs/$route_photo");
			
		if ($new_width && ($width_orig < $height_orig)) {
				$new_width = ($new_height / $height_orig) * $width_orig;
			} else {
				$new_height = ($new_width / $width_orig) * $height_orig;
				}
	
	//resample the image
		$thumb = ImageCreateTrueColor($new_width, $new_height);
		$tmp_image = ImageCreateFromJPEG("../images/climbs/$route_photo");
		ImageCopyResampled($thumb, $tmp_image, 0, 0, 0, 0, $new_width, 
$new_height, $width_orig, $height_orig);
	
		ImageJPEG($thumb, $route_photo, '100');
		ImageDestroy($thumb);
	}
?>

//Image Tag:
echo ("<img src='include/function_gd.php?route_photo=$thumb'  
align='center' id='image'>");

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