Question

我正在尝试打开存储在本地存储中的文件。但是每次我单击“打开”时，都会出现此错误403。我认为这是在我尝试添加删除按钮并用我的php代码创建并发症时发生的。

请检查此代码：

<?php
$con=mysqli_connect("localhost","root","","annualdb");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$result = mysqli_query($con,"SELECT * FROM asean_japan WHERE agency='Rail'");

?>
<table class='page'>
  <tr>
    <th>Select</th>
    <th>Agency</th>
    <th>FileName</th>
    <th>FileType</th>
    <th>Date Received</th>
    <th>Action</th>
  </tr>
<?php
while($row = mysqli_fetch_array($result))
{
  ?>
  <tr>
    <td align='center' bgcolor='#FFFFFF'><input name='checkbox[]' type='checkbox' value='<?php echo $row['id']; ?>'></td>
    <td><?php echo $row['agency']; ?></td>
    <td><?php echo $row['filename']; ?></td>
    <td><?php echo $row['filetype']; ?></td>
    <td><?php echo $row['date']; ?></td>
    <td><a target='_blank' href='../annual/indicators/" <?php echo $row['filename']; ?>"'>OPEN</a></td>
  </tr>

<?php
}
?>
</table>
<?php
mysqli_close($con);
?>

Answer 1

我尝试将表格放在php标记内，但突然起作用。这可能是什么问题？

echo "<table class='page'>
<tr>
<th>ID</th>
<th>Agency</th>
<th>FileName</th>
<th>FileType</th>
<th>Date Received</th>
<th>Action</th>
</tr>";

while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['id'] . "</td>";
echo "<td>" . $row['agency'] . "</td>";
echo "<td>" . $row['filename'] . "</td>";
echo "<td>" . $row['filetype'] . "</td>";
echo "<td>" . $row['date'] . "</td>";
echo "<td><a target='_blank' href='../annual/CPA/" . $row['filename'] . "'>OPEN</a></td>";
echo "</tr>";
}
echo "</table>";

mysqli_close($con);
?>

Answer 2

<td><a target='_blank' href='../annual/indicators/" <?php echo $row['filename']; ?>"'>OPEN</a></td>

该行无效。它将创建如下链接： http://myhost.com/annual/indicators/%22%20somefile.pdf%22，其中％22是“的urlencoded值，％20是空白的urlencoded值。

您可以通过以下方式使用它：

<td><a target='_blank' href='../annual/indicators/<?php echo $row['filename']; ?>'>OPEN</a></td>

错误403禁止访问！您无权访问所请求的对象。服务器对其进行了读保护或不可读

2 个答案: