在这里的代码中，$fetch['id']工作良好的其他地方，但在窗口。打开它总是取相同的值，那就是id号25，而不是其他id，我不明白为什么会这样....请帮助 ........
$i=1;
while ($fetch = mysql_fetch_array($data_qry))
{
echo $url = $_SERVER['REQUEST_URI']."/edit_user_comment.php?id=".$fetch['id']; // here variable working fine....
echo "<tr>";
echo "<td>".$i."</td>";
echo "<td>".$fetch['comp_name']."</td>";
if (strlen($fetch['comp_add'])> 10)
{
echo "<td>".substr($fetch['comp_add'], 0, 10)."...";
}else{
echo "<td>".$fetch['comp_add'];
}"</td>";
echo "<td>".$fetch['business_registration_no']."</td>";
echo "<td>".$fetch['email_id']."</td>";
echo "<td>".$fetch['tel_no']."</td>";
echo "<td>".$fetch['fax']."</td>";
echo "<td>".$fetch['prsn_in_chrg']."</td>";
echo "<td>".$fetch['ph_no']."</td>";
?>
<script>
function windowopen(){
window.open('<?= $url; ?>','Comment Edit', 'menubar=1,location=1,status=1,scrollbars=yes,width=500, height=500');
//here it is takin $fetch['id']=25 only and not other ids, i want to know the mistake i m doing here... Please help
}
</script>
<?php
if($fetch['comments']!='')
{
echo "<td><textarea name='comments' id='id_comments'>".$fetch['comments']."</textarea>
<a onclick='return windowopen();' class='button button-primary button-large edit'>Edit</a></td>";
}else{
echo "<td><a onclick='return windowopen();' class='button button-primary button-large'>Add Comment</a></td>";
}
echo "<td> <a href='response.php?id=".$fetch['id']."&action=approve' class='button button-primary button-large approve'>Approve</a> <a href='response.php?id=".$fetch['id']."&action=dissapprove' class='button button-primary button-large'>Dissapprove</a> </td>";
echo "</tr>";
$i++;
}
详细说明我之前的评论-生成的链接应该看起来像:
echo '<td><a onclick="return windowopen('''.$url.''');">Add Comment</a></td>';
…你的函数应该在循环外声明一次，像这样:
<script>
function windowopen(url)
{
window.open(url,'Comment Edit', '/*blah*/');
}
</script>
这是因为您重新声明了相同的函数windowopen()所以最后的声明覆盖了其他的并且您得到了所有元素的相同函数