php代码中的弹出窗口

Question

I'm trying to do table where pressing on ID, it opens popup window, where will be menu, for example, change row values, add comments.. But I have a problem with generating link for each row:

foreach ($results as $row) {
// Each $row is a row from the query
$rowid = $row->ID;
echo '<tr><td>';
echo '<a href="#" onclick="javascript:window.open("http://192.168.210.140/todolist/controls-menu?funnelid="'.$rowid.',"Controls Menu","width= 700,height= 500,toolbar= no,location= no,directories= 0,status= no,menuBar= no,scrollBars= no,resizable= yes,left= 400,top= 150,screenX= 400,screenY= 150");">'.$rowid.'</a>';

Maybe it's need to add

<script type="text/javascript>
... Some script when pressing <a href="#">
</script>

But I have no idea, how to detect from script which link is pressed, for example is it

http://192.168.210.140/todolist/controls-menu?funnelid=999 

OR

http://192.168.210.140/todolist/controls-menu?funnelid=1100

PS I'm using Wordpress.

Answer 1

HTML Code

<?php  $id=1;?>
<a href="" onClick="popitup('popup1.php?id=<?php echo $id;?>')">Open</a>

Javascript Code

<script type="text/javascript">
function popitup(url) {
newwindow=window.open(url,'name','height=200,width=150');
if (window.focus) {newwindow.focus()}
return false;
}
</script>