<?php
$dbhost = 'xxxx';
$dbuser = 'xxxx';
$dbpass = 'xxxx';
$dbname = 'xxxx';
$conn = mysql_connect($dbhost, $dbuser, $dbpass) or die ('Error connecting to mysql');
mysql_select_db($dbname);
$result = mysql_query("SELECT * FROM mytable");
$row = mysql_fetch_array($result)
?>
<?php foreach ($rows as $row): ?>
<tr align="center">
<td><?php echo htmlspecialchars($row['Picturedata']); ?></td>
</tr>
<?php endforeach; ?>
我得到一个错误: 警告:()提供的foreach无效参数我试图做一个ASP像中继器在PHP
这工作 Thans! – Joe 2009-10-16 21:41:30