hi, my first post here

so i am having a problem in this line of code...

    $display = imagecreatefromjpeg("$dir/$image");  // Read all the images into an array.

in the display cell it doesnt display the image. it gives me "Resource id #12".

the file name and file size cells display the correct info and the javascript displays the correct image...

what am i doing wrong?

i am using php 5+ and gd 2+ is enabled.

thanks. full code follows.

<div align="center">Click on an image to view it in a separate window.</div><br />
<table align="center" cellspacing="5" cellpadding="5" border="1">
   <tr>
      <td align="center"><b>Image</b></td>
      <td align="center"><b>Image Name</b></td>
      <td align="center"><b>Image Size</b></td>
   </tr>
<?php # images.php
// This script lists the images in the uploads directory.

$dir = 'uploads'; // Define the directory to view.

$files = scandir($dir); // Read all the images into an array.

// Display each image caption as a link to the JavaScript function.
foreach ($files as $image) {

   if (substr($image, 0, 1) != '.') { // Ignore anything starting with a period.

      // Get the image's size in pixels.
      $image_size = getimagesize("$dir/$image");

      // Calculate the image's size in kilobytes.
      $file_size = round ( (filesize ("$dir/$image")) / 1024) . "kb";

    // display gif
    $display = imagecreatefromjpeg("$dir/$image"); // Read all the images into an array.

      // Print the information.
      echo "<tr>
         <td>$display</td>
         <td><a href=\"javascript: create_window('$dir/$image',$image_size[0],$image_size[1])\">$image</a></td>
         <td>$file_size</td>
      </tr>";

   } // End of the IF.

} // End of the foreach loop.

?>
</table>

so it turns out i was missing some simple html...

         <td><img src='$dir/$image' /></td>

i didnt need to use $display in the td.

all working now.