我试图用PHP中的转发器做ASP

时间:2009-10-16 21:00:13

标签: php mysql

<?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; ?>

我收到错误: 警告:为foreach()提供的参数无效

2 个答案:

答案 0 :(得分:0)

它应该是$ rows而不是$ row

$rows = mysql_fetch_array($result)

答案 1 :(得分:0)

您需要执行while循环来拉取每个行集。 mysql_fetch_array()一次只能拉一行。考虑这个解决方案:

<?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, $conn);
$result = mysql_query("SELECT * FROM mytable", $conn);

while ($row = mysql_fetch_array($result)) {
    echo '<tr align="center"><td>' . htmlspecialchars($row['Picturedata']) . '</td></tr>';
}
?>