my home.php是
required("conn.php");
<html>
<body>
<form action="like.php" name="art" method="post">
<?php
//$email=$data['email'];
?>
<table align="center" border="2">
<tr><th colspan="4"><h2 align="center">Articals</h2></th></tr>
<tr> <th>Name</th><th>Artical</th><th>Status</th><th colspan="2">option</th>
</tr>
<?php
$r = mysql_query("select * from ar ");
while($ro=mysql_fetch_array($r)){
?>
<tr>
<td> <?php echo $ro["fname"]?></td>
<td><textarea rows="2" cols="25"><?php echo $ro["post"]?></textarea></td>
<td><?php echo $ro["status"]?></td>
<td><a name="like" href="like.php?id=<?php echo $ro['id']?>"> like</a></td>
<!--<td><a href="delart.php?id=<?php //echo $row["id"]?>" >Delete</a></td>
<td><a href="edt.php?unm=<?php //echo $row[0]?>">Upadate</a></td>
-->
</tr>
<?php
}
mysql_close($con);
?>
</form>
</table>
</body>
</html>
php就是
<?php
session_start();
if(!isset($_SESSION['user']))
{
header('Location:index.php');
}
require("conn.php");
$id=$_GET['id'];
mysql_query("insert into `like` ar where id='$id'");
?>
当我点击这样的链接时,错误"在这台服务器上找不到请求的URL/25-8/like.php。"来了,请帮助我是php的新手我想在那里计算没有点赞,点击点赞按钮后,它会自动变为不同
YOu必须将home.php和like.php放在同一目录中吗?