我没有错误,连接正常,但我的数据库没有收到任何数据。我一直在关注一个教程而不是我自己的代码,但是我无法看到我在哪里犯了错误。
我创建了一个名为files的表,它与下面代码中使用的参数相对应。
代码位于名为uploader.php的文件中:
<?php if(isset($_POST['uploader']) && $_FILES['userfile']['size'] > 0)
{
echo "working";
$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);
}
mysql_connect("localhost","root","root");
mysql_select_db("globe_bank");
$query = "INSERT INTO files (name, size, type, content ) ".
"VALUES ('$fileName', '$fileSize', '$fileType', '$content')";
mysql_query($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="uploader" type="submit" name="uploader" value=" Uploader " /></td>
</tr>
</tbody>
</table>
</form>