9

我正在用 Python ftplib 开发一个 FTP 客户端。如何为其添加代理支持(我见过的大多数 FTP 应用程序似乎都有它)?我特别在考虑 SOCKS 代理,还有其他类型... FTP、HTTP(甚至可以将 HTTP 代理与 FTP 程序一起使用吗?)

任何想法如何做到这一点?

4

6 回答 6

8

根据这个来源。

取决于代理,但常用的方法是 ftp 到代理,然后使用目标服务器的用户名和密码。

例如对于 ftp.example.com:

Server address: proxyserver (or open proxyserver from with ftp)
User:           anonymous@ftp.example.com
Password:       password

在 Python 代码中:

from ftplib import FTP
site = FTP('my_proxy')
site.set_debuglevel(1)
msg = site.login('anonymous@ftp.example.com', 'password')
site.cwd('/pub')
于 2009-08-18T12:44:33.820 回答
4

您可以.urllib2

ph = urllib2.ProxyHandler( { 'ftp' : proxy_server_url } )
server= urllib2.build_opener( ph )
于 2009-08-18T12:52:04.160 回答
3

我遇到了同样的问题,需要使用ftplib模块(不要用 URLlib2 重写我的所有脚本)。

我设法编写了一个脚本,在套接字层(由 ftplib 使用)上安装透明HTTP 隧道。

现在,我可以通过 HTTP透明地进行 FTP 了!

你可以在那里得到它: http ://code.activestate.com/recipes/577643-transparent-http-tunnel-for-python-sockets-to-be-u/

于 2011-04-07T13:31:39.223 回答
2

标准模块ftplib不支持代理。似乎唯一的解决方案是编写您自己的自定义版本的ftplib.

于 2009-08-18T12:42:28.013 回答
2

修补内置套接字库绝对不是每个人的选择,但我的解决方案是socket.create_connection()在主机名匹配白名单时修补以使用 HTTP 代理:

from base64 import b64encode
from functools import wraps
import socket

_real_create_connection = socket.create_connection
_proxied_hostnames = {}  # hostname: (proxy_host, proxy_port, proxy_auth)


def register_proxy (host, proxy_host, proxy_port, proxy_username=None, proxy_password=None):
    proxy_auth = None
    if proxy_username is not None or proxy_password is not None:
        proxy_auth = b64encode('{}:{}'.format(proxy_username or '', proxy_password or ''))
    _proxied_hostnames[host] = (proxy_host, proxy_port, proxy_auth)


@wraps(_real_create_connection)
def create_connection (address, *args, **kwds):
    host, port = address
    if host not in _proxied_hostnames:
        return _real_create_connection(address, *args, **kwds)

    proxy_host, proxy_port, proxy_auth = _proxied_hostnames[host]
    conn = _real_create_connection((proxy_host, proxy_port), *args, **kwds)
    try:
        conn.send('CONNECT {host}:{port} HTTP/1.1\r\nHost: {host}:{port}\r\n{auth_header}\r\n'.format(
            host=host, port=port,
            auth_header=('Proxy-Authorization: basic {}\r\n'.format(proxy_auth) if proxy_auth else '')
        ))
        response = ''
        while not response.endswith('\r\n\r\n'):
            response += conn.recv(4096)
        if response.split()[1] != '200':
            raise socket.error('CONNECT failed: {}'.format(response.strip()))
    except socket.error:
        conn.close()
        raise

    return conn


socket.create_connection = create_connection

我还必须创建一个忽略host返回的PASVEPSVFTP 命令的 ftplib.FTP 子类。示例用法:

from ftplib import FTP
import paramiko  # For SFTP
from proxied_socket import register_proxy

class FTPIgnoreHost (FTP):
    def makepasv (self):
        # Ignore the host returned by PASV or EPSV commands (only use the port).
        return self.host, FTP.makepasv(self)[1]

register_proxy('ftp.example.com', 'proxy.example.com', 3128, 'proxy_username', 'proxy_password')

ftp_connection = FTP('ftp.example.com', 'ftp_username', 'ftp_password')

ssh = paramiko.SSHClient()
ssh.set_missing_host_key_policy(paramiko.AutoAddPolicy())  # If you don't care about security.
ssh.connect('ftp.example.com', username='sftp_username', password='sftp_password')
sftp_connection = ssh.open_sftp()
于 2016-05-10T16:28:06.777 回答
0

这是使用requests不支持 CONNECT 隧道的 squid 代理测试的解决方法:

def ftp_fetch_file_through_http_proxy(host, user, password, remote_filepath, http_proxy, output_filepath):
    """
    This function let us to make a FTP RETR query through a HTTP proxy that does NOT support CONNECT tunneling.
    It is equivalent to: curl -x $HTTP_PROXY --user $USER:$PASSWORD ftp://$FTP_HOST/path/to/file
    It returns the 'Last-Modified' HTTP header value from the response.

    More precisely, this function sends the following HTTP request to $HTTP_PROXY:
        GET ftp://$USER:$PASSWORD@$FTP_HOST/path/to/file HTTP/1.1
    Note that in doing so, the host in the request line does NOT match the host we send this packet to.

    Python `requests` lib does not let us easily "cheat" like this.
    In order to achieve what we want, we need:
    - to mock urllib3.poolmanager.parse_url so that it returns a (host,port) pair indicating to send the request to the proxy
    - to register a connection adapter to the 'ftp://' prefix. This is basically a HTTP adapter but it uses the FULL url of
    the resource to build the request line, instead of only its relative path.
    """
    url = 'ftp://{}:{}@{}/{}'.format(user, password, host, remote_filepath)
    proxy_host, proxy_port = http_proxy.split(':')

    def parse_url_mock(url):
        return requests.packages.urllib3.util.url.parse_url(url)._replace(host=proxy_host, port=proxy_port, scheme='http')

    with open(output_filepath, 'w+b') as output_file, patch('requests.packages.urllib3.poolmanager.parse_url', new=parse_url_mock):
        session = requests.session()
        session.mount('ftp://', FTPWrappedInFTPAdapter())
        response = session.get(url)
        response.raise_for_status()
        output_file.write(response.content)
        return response.headers['last-modified']


class FTPWrappedInFTPAdapter(requests.adapters.HTTPAdapter):
    def request_url(self, request, _):
        return request.url
于 2016-09-02T14:44:17.783 回答