2012-01-05 89 views
-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'; 
?> 
+1

这是一个很棒的故事。 – 2012-01-05 16:35:21

+5

**不要将用户输入放入SQL查询中**多少次...... – 2012-01-05 16:35:50

回答

3

echo "</tr>"; IM去你的while循环中。

+0

仍然不起作用表格的标题显示在网页上,但来自数据库aint的数据显示在网页上其blanc – user1130533 2012-01-05 16:43:23