Question

我有一个回显按钮的表，目前这个按钮将你带到'details.php'，并将£booking_id记录转移到新的php scrpit。这是我目前的代码：

echo '<td>a href="Details.php?id='.$row['booking_id'].'"><button>View Details</button></td>';

我想要这个按钮打开一个新的弹出窗口。我如何将两者合并在一起，所以当我点击按钮弹出窗口并将'booking_id'记录结束时。我有java脚本代码：

<script type="text/javascript">
// Popup window code
function newPopup(url) {
    popupWindow = window.open(
        url,'popUpWindow','height=700,width=800,left=10,top=10,resizable=yes,scrollbars=yes,toolbar=yes,menubar=no,location=no,directories=no,status=yes')
}
</script>
<a href="JavaScript:newPopup('Details.php);">View Details</a>

Answer 1

echo '<td>';
echo '<input type="button" value="View Details" onclick="JavaScript:newPopup(\'Details.php?id='.$row['booking_id'].'\')" />';
echo '</td>';

Answer 2

<强>的index.php

<html>
<head><title>No Head</title></head>
<body>    
<?php    
$rows = array(
    0 => array('booking_id' => 1),
    1 => array('booking_id' => 2),
    2 => array('booking_id' => 3),
    3 => array('booking_id' => 4),
);    
echo '<table>';
foreach($rows as $row)
{
    echo "\n<tr><td>";
    echo '<a href="Details.php?id=' . $row['booking_id'] .'"';
    echo ' onclick="javascript:popup(this.href); return false;">';
    echo 'View Details';
    echo '</a>';
    echo "</td></tr>";
}
echo "\n</table>";
?>    
</body>
<script type="text/javascript">
function popup(url) {
    popupWindow = window.open( url, 'popUpWindow',
        "height=700,width=800,left=10,top=10,resizable=yes,scrollbars=yes,toolbar=yes,menubar=no,location=no,directories=no,status=yes"
    )
}
</script>
</html>

<强> DETAILS.PHP

<html>
<head></head>
<body>
Details for ID <?php echo $_GET['id']; ?>
</body>
</html>

PHP JavaScript：PopUp窗口

2 个答案: