您好我正在尝试制作一个上传然后下载该文件的应用程序，但我在打开除txt文件之外的文件时遇到问题。我收到一条错误消息，表示＆＃34;未能打开＆＃34;。
请帮我解决这个问题所以我提供了我的代码。
谢谢。
所以这是我的代码：
upload.php的
<?php if(isset($_POST['upload']) && $_FILES['userfile']['size'] > 0)
{
$fileName = $_FILES['userfile']['name'];
$tmpName = $_FILES['userfile']['tmp_name'];
fileSize = $_FILES['userfile']['size'];
$fileType = $_FILES['userfile']['type'];
$fp = fopen($tmpName, 'r');
$content = fread($fp, filesize($tmpName));
$content = addslashes($content);
fclose($fp);
if(!get_magic_quotes_gpc())
{
$fileName = addslashes($fileName);
}
include_once 'dbConnect.php';
$query = "INSERT INTO gravator (name, size, type, content ) ".
"VALUES ('$fileName', '$fileSize', '$fileType', '$content')";
mysqli_query($conn,$query) or die('Error, query failed');
echo " File $fileName uploaded";
}
?>
<form method="post" enctype="multipart/form-data">
<table width="350" border="0" cellspacing="1" cellpadding="1">
<tbody>
<tr>
<td width="246">
<input type="hidden" name="MAX_FILE_SIZE" value="2000000" />
<input id="userfile" type="file" name="userfile" /></td>
<td width="80"><input id="upload" type="submit" name="upload" value="Upload
" /></td>
</tr>
</tbody>
</table>
</form>
upload.php的
<?php
include_once 'dbConnect.php';
$sql = "Select * from gravator";
$res = mysqli_query($conn,$sql) or die('Error, query failed');
if(isset($_GET['id'])) { // if id is set then get the file with the id from
database
$id = $_GET['id'];
$query = "SELECT name, type, size, content FROM gravator WHERE id = $id";
$result = mysqli_query($conn,$query) or die('Error, query failed');
list($name, $type, $size, $content) =
mysqli_fetch_array($result);
header("Content-length: $size");
header("Content-type: $type");
header("Content-Disposition: attachment; filename=$name");
echo $content; exit;
}
?>
<?php
$query = "SELECT id, name FROM gravator";
$result = mysqli_query($conn,$query) or die('Error, query failed');
if(mysqli_num_rows($result) == 0)
{
echo "Database is empty";
}
else
{
while(list($id, $name) = mysqli_fetch_array($result))
{
?>
<a href="download.php?id=<?php echo $id;?>"><?php echo $name; ?></a>
<?php
}
}
?>