-3
IM试图从一个名为“议程”表中的所有信息显示在我的网页时,它已经批准,但有麻烦信息没有显示
<?php
$agenda_id = $_GET['agenda_id'];
include 'library/connect.php';
$result = mysql_query("SELECT * FROM agenda WHERE approval = 'approved' AND agenda_id = '$agenda_id'");
echo "<table border='1'><tr><th>Subject</th><th>Duration</th></tr>";
while($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['subject']. "</td>";
echo "<td>" . $row['duration']. "</td>";
}
echo "</tr>";
echo "</table>";
include 'library/closedb.php';
?>
这是一个很棒的故事。 – 2012-01-05 16:35:21
**不要将用户输入放入SQL查询中**多少次...... – 2012-01-05 16:35:50