看下面两句话：
$path_parts = pathinfo($filename);
$title = $path_parts['filename'];在如果文件名是GBK、GB2312等非UTF-8或ASCI字符集编码时，获取的filename属性是错误的。
因此，如果要获取争取的属性，应当先对字符集进行编码。
$filename = transToUTF8($filename);
$path_parts = pathinfo($filename);
$title = $path_parts['filename'];
