Passing variable to popup

[biscutty]

Hi, I am making a script that displays a directory of thumbnail images in a table.

I want to make the thumbnalis link to a new php file in a popup window. This file then needs to know which image to open, corresponding to the dynamicly listed thumbnail.

I want to be able to spawn multiple pages (open new pages for each thumbnail) so more than one image can be opened at a time.

I understand how to pass the variable of the image name to the new window but need to controll its size to match the image size.

Any help would be appreciated.

thanks Biscutty

my script so far:

<table width="440" height="20" cellpadding="3" cellspacing="10" class="contactsheet" >
  <tr> 

  <?  	

$dirName = "Portfolio/Adult/thumbs"; 
$dir = opendir("$dirName"); 

$count = 0; 

	while ($file = readdir($dir)){ 
		if ( eregi(".jpg$","$file")){ 
		$count++; 

		if ($count == 1){ print ("<tr>\n"); } 

		print ("<td><img src=\"$dirName/$file\">\n");
		print ("</td>");


		if ($count >= 3){ print ("</tr>\n"); $count = 0; } 
		} 
	}

	?>
</tr>
  </table>

:rolleyes:

[Pig]

Just make your url's for the links carry the id in the query string:
http://www.domain.com/page.php?id=$dirName/$file

Then just have the src on the called page look for the id.
if( empty($GET['id']) ) echo("You must specify an image");
else echo("<img src=\"" . $GET['id'] . "\">");