Javascript弹出窗口没有从php父页面获取值

Question

What I'm doing wrong?

This is the JavaScript code:

    <script language="javascript">
    function popupIp(url) 
    {
    window.open(url,'popupWindow','toolbar=no,location=no,directories=no,status=no,
menubar=no,scrollbars=no,resizable=no,copyhistory=no,width=350,height=320,screenX=150,
screenY=150,top=50,left=50')
    }
    </script>
    </head>

And here I call the popup:

    <a href="#" <?='onclick="popupIp(\''.tep_href_link('ip-address-popup.php',  'ut='
 . $visit['bp_ip']), '&submit=Process' .'\'); return false;"';?>>
    <?php echo $visit["bp_ip"] . '</a>

I echo $visit["bp_ip"] and $ut on popup but there are no values.

Answer 1

<?php
$popup = '<a href="#" onclick="popupIp(\'';
$popup .= tep_href_link('ip-address-popup.php', 'ut=' . $visit['bp_ip']);
$popup .= '&submit=Process\'); return false;">';
$popup .= $visit["bp_ip"];
$popup .= '</a>';
echo $popup;
?>

Your example had so many errors in it, hidden by putting it all into one unreadable line, that it took a while to unravel it. One liners ALWAYS hide bugs and make code hard to read, it's a habit well worth breaking if you plan on doing more programming, particularly on a professional level. Clean, debuggable, easy to read, code, that's the key.

I'm not 100% sure this is what you intended, but it's at least readable and debuggable so you can figure out what is happening.