问题描述

我正在尝试将文件上传到 FTP.

I'm trying to upload a file to FTP.

这是我的代码:

$connect = ftp_connect('ftp.my-server.fr');
$login = ftp_login($connect, 'username', 'pass');

$remote_file = '/' . $date;
$local_file = fopen('C:/MAMP/htdocs/mysite/myfolder/' . $hour .'.mp3', 'r');

ftp_chdir($connect, '/'.$date);

if (ftp_fput($connect, $remote_file, $local_file, FTP_ASCII)) {
    echo "The file $local_file has been loaded";
} else {
    echo "Error while uploading file " . $local_file;
}

我收到一个错误:

PHP 警告:ftp_fput():无法打开该文件:是第 26 行 C:MAMPhtdocsmysiteindex.php 中的目录

PHP Warning: ftp_fput(): Can't open that file: Is a directory in C:MAMPhtdocsmysiteindex.php on line 26

我不明白，因为路径是文件.当我在浏览器中粘贴 $local_file URL 时，声音正在播放.

I don't understand because the path is the file. When I paste $local_file URL in my browser the sound is playing.

推荐答案

你的 $local_file 是可以的，但是你的 $remote_file 是一个目录(你使用 '/' . $date for ftp_chdir)，它需要是一个文件的路径(将被创建)

Your $local_file is OK, but your $remote_file is a directory (you use '/' . $date for ftp_chdir), and it need to be a path to a file (that will be created)

您可以使用 basename 复制与本地文件相同的文件名:

You can copy the same filename than the local file with basename :

$remote_dir = '/' . $date;
$local_file = fopen('C:/MAMP/htdocs/mysite/myfolder/' . $hour .'.mp3', 'r');

ftp_chdir($connect, $remote_dir);
$remote_file = $remote_dir . '/' . basename($local_file) ;

if (ftp_fput($connect, $remote_file, $local_file, FTP_ASCII)) {
    echo "The file $local_file has been loaded";
} else {
     echo "Error while uploading file " . $local_file;
}

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