[Resolved] pass variable to popup inside for loop

beginphp04

hello
using weedpacket's code:

$sql = "SELECT employee_photo, employee_name, employee_id FROM employees WHERE name_id = '$name_id'";

$result = mysql_query($sql) or die (mysql_error());

$rows = mysql_num_rows($result);

if($rows > 0) {

$nowbreak = 4;  

echo "<table>\n";  
echo "<tr>\n";  

for($i=1; $i<=$rows; $i++) { 
    $row=mysql_fetch_array($result); 
    echo "<td><img src=\"images/".$row['employee_photo']."\"><br />".$employee_name."</td>\n";  

    if($i%$nowbreak==0) {  
        echo "</tr><tr>\n";  
    }  
}  
echo "</tr></table>";

}

what is the best way to go about passing the employee_id variable to a popup? so each employee photo will be <a href=... including the employee_id variable.

there are lots of examples around (i have spent the last few hours looking) but all where the javascript is outside the php then they just use
<?php echo '$row['employee_id']' ?>
in the appropriate place. i cannot seem to do that because of the loop. i would appreciate any tips.
many thanks

beginphp04

i think i am almost (but not quite) there with this:

// then

it is passing the correct id variable to the correct image when i hover over the image, but it will not launch any popup.

sorry about not using the proper

 layout only i couldn't figure out how.  which probably says a lot about why i am struggling with this code 

:)

beginphp04

solution for anyone searching this thread in future:

<SCRIPT LANGUAGE="JavaScript1.2">
function openWindow(theURL,winName,features)
{
window.open(theURL,winName,features);
}
</SCRIPT>

// then