0
<tbody>
<?php
$query = mysql_query("select * from announce ") or die(mysql_error());
while ($row = mysql_fetch_array($query)) {
$date = $row['date'];
$subject = $row['subject'];
$ann = $row['ann'];
$sender = $row['Sender'];
$id = $row['id'];
?>
<tr class="odd gradeX">
<td><?php echo $row['date']; ?></td>
<td><?php echo $row['subject']; ?></td>
<td><?php echo $row['ann']; ?></td>
<td><?php echo $row['Sender']; ?></td>
<td><a class='inline' href="#inline_content" id="<?php echo $row['id'] ?>"> Click Me</a></td>
</tr>
<div style='display:none'>
<div id='inline_content' style='padding:10px; background:#fff;'>
<p><strong><?php echo $row['subject']; ?></strong></p> <p align="right"><?php echo $row['date']; ?></p>
<p?><?php echo $row['ann']; ?>.</p>
<br><br>enter code here
<p>By: <?php echo $row['Sender']; ?> </p> </div>
<?php } ?>
</tbody>
我彈出一個jQuery的燈箱,但燈箱中的信息是在數據庫中,當一個按鈕調用它會彈出的信息,當我點擊另一個按鈕它有第一個相同的記錄,但我打電話給別人嗎?請需要幫助!提前致謝!使用php在MySQL中列出記錄,但只有一條記錄出現在所有?