//index.php
<html>
<head>
<title>新闻系统</title>
</head>
<body>
<table border="1">
<tr>
<?php
include("conn.php");
$sql="select * from `category` order by `order_id` desc";
$rs=mysql_query($sql);
while($row=mysql_fetch_array($rs))


?>
<td><a href="list.php?category=<?php echo $row[order_id]?>"><?php echo $row[category_name]?></a></td>
<?php
}
?>
</tr>
</table>
</body>
</html>//list.php
<?php
include("head.php")
$id=$_GET['category'];
?>
<html>
<head>
<title>新闻系统</title>
</head>
<body>
<table border="1">
<tr>
<td></td>
</tr>
<tr>
<td></td>
</tr>
<?php
include("conn.php");
$sql="select * from `news` where `order_id`='$id'";
$rs1=mysql_query($sql);
while($row1=mysql_fetch_array($rs1))
{
?>
<tr>
<td><a href="display.php?id=<?php echo $row[id] ?>"><?php echo $row1[title]?></a>
</td>
</tr>
<?php
}
?></table>
</body>
</html>想让list.php获取index.php中category中的ID,再在display.php中按category的不同显示,但是运行一下结果是
Parse error: syntax error, unexpected T_VARIABLE in F:\www\mynews\list.php on line 3
请各位高手相助!