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我有一个网站需要连接到位于专用网络防火墙后面的 SOAP API。我已经尝试了一些不同程度的成功。

1.SSH隧道

我尝试使用以下方法在网站上设置 SSH 隧道（IP 是随机示例）。

export WEB_HOST=203.0.113.10
export LINUX_HOST_WITH_PUBLIC_IP_ON_PRIVATE_NETWORK=203.0.113.11
export SOAP_API_HOST=198.51.100.10

# from $LINUX_HOST_WITH_PUBLIC_IP_ON_PRIVATE_NETWORK I run the following
ssh -f $LINUX_HOST_WITH_PUBLIC_IP_ON_PRIVATE_NETWORK -L 4000:$SOAP_API_HOST:80 -N

# test with curl
$ curl -I http://localhost:4000
HTTP/1.1 200 OK
...

但是，当我尝试将 Suds 与 API 一起使用时，它似乎不起作用。

$ cat temp.py

from suds.client import Client

url = 'http://localhost:4000/scripts/WebObjects.exe/WebServices.woa/ws/Law?wsdl'
client = Client(url)
print(client.service.doThing())

$ python temp.py
Traceback (most recent call last):
  File "temp.py", line 6, in <module>
    print(client.service.doThing())
  File "/Users/foouser/.virtualenvs/project/lib/python2.7/site-packages/suds/client.py", line 542, in __call__
    return client.invoke(args, kwargs)
  File "/Users/foouser/.virtualenvs/project/lib/python2.7/site-packages/suds/client.py", line 602, in invoke
    result = self.send(soapenv)
  File "/Users/foouser/.virtualenvs/project/lib/python2.7/site-packages/suds/client.py", line 637, in send
    reply = transport.send(request)
  File "/Users/foouser/.virtualenvs/project/lib/python2.7/site-packages/suds/transport/https.py", line 64, in send
    return  HttpTransport.send(self, request)
  File "/Users/foouser/.virtualenvs/project/lib/python2.7/site-packages/suds/transport/http.py", line 77, in send
    fp = self.u2open(u2request)
  File "/Users/foouser/.virtualenvs/project/lib/python2.7/site-packages/suds/transport/http.py", line 118, in u2open
    return url.open(u2request, timeout=tm)
  File "/usr/local/Cellar/python/2.7.5/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 404, in open
    response = self._open(req, data)
  File "/usr/local/Cellar/python/2.7.5/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 422, in _open
    '_open', req)
  File "/usr/local/Cellar/python/2.7.5/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 382, in _call_chain
    result = func(*args)
  File "/usr/local/Cellar/python/2.7.5/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 1214, in http_open
    return self.do_open(httplib.HTTPConnection, req)
  File "/usr/local/Cellar/python/2.7.5/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 1184, in do_open
    raise URLError(err)
urllib2.URLError: <urlopen error [Errno 61] Connection refused>

2. 使用 python contextmanager 包装 API 调用

我还尝试使用基于Fabric代码的 python contextmanager 包装 API 调用。

$ cat temp2.py                                                                                                     
from contextlib import contextmanager
import socket
import paramiko
import logging


@contextmanager
def remote_tunnel(remote_port, local_port=None, local_host="localhost", remote_bind_address="127.0.0.1", transport=None):
    if local_port is None:
        local_port = remote_port

    sockets = []
    channels = []
    threads = []

    def accept(channel, (src_addr, src_port), (dest_addr, dest_port)):
        channels.append(channel)
        sock = socket.socket()
        sockets.append(sock)

        try:
            sock.connect((local_host, local_port))
        except Exception, e:
            print "[%s] rtunnel: cannot connect to %s:%d (from local)" % (env.host_string, local_host, local_port)
            chan.close()
            return

        print "[%s] rtunnel: opened reverse tunnel: %r -> %r -> %r"\
              % (env.host_string, channel.origin_addr,
                 channel.getpeername(), (local_host, local_port))

        th = ThreadHandler('fwd', _forwarder, channel, sock)
        threads.append(th)

    transport.request_port_forward(remote_bind_address, remote_port, handler=accept)

    try:
        yield
    finally:
        for sock, chan, th in zip(sockets, channels, threads):
            sock.close()
            chan.close()
            th.thread.join()
            th.raise_if_needed()
        transport.cancel_port_forward(remote_bind_address, remote_port)


def main():

    WEB_HOST = '203.0.113.10'
    LINUX_HOST_WITH_PUBLIC_IP_ON_PRIVATE_NETWORK = '203.0.113.11'
    SOAP_API_HOST = '198.51.100.10'
    LOCAL_PORT = 4000
    REMOTE_PORT = 80
    SSH_USER = 'foouser'    

    # Connect to SSH host
    client = paramiko.SSHClient()
    client.load_system_host_keys()
    client.set_missing_host_key_policy(paramiko.WarningPolicy())
    ssh_host = (LINUX_HOST_WITH_PUBLIC_IP_ON_PRIVATE_NETWORK, 22, SSH_USER)

    logging.debug('Connecting to ssh host {}:{:d} ...'.format(ssh_host[0], ssh_host[1]))
    try:
        client.connect(ssh_host[0], ssh_host[1], username=ssh_host[2], key_filename=None, look_for_keys=True, password=None)
    except Exception as e:
        logging.error('Failed to connect to {}:{:d}: {:r}' % (ssh_host[0], ssh_host[1], e))

    with remote_tunnel(remote_port=REMOTE_PORT, local_port=LOCAL_PORT, local_host='localhost', remote_bind_address=SOAP_API_HOST, transport=client.get_transport()):
        print(requests.get('http://localhost:4000/'))

if __name__ == '__main__':
    main()


$ python temp2.py
Traceback (most recent call last):
  File "temp2.py", line 80, in <module>
    main()
  File "temp2.py", line 76, in main
    with remote_tunnel(remote_port=REMOTE_PORT, local_port=LOCAL_PORT, local_host='localhost', remote_bind_address=SOAP_API_HOST, transport=client.get_transport()):
  File "/usr/local/Cellar/python/2.7.5/Frameworks/Python.framework/Versions/2.7/lib/python2.7/contextlib.py", line 17, in __enter__
    return self.gen.next()
  File "temp2.py", line 35, in remote_tunnel
    transport.request_port_forward(remote_bind_address, remote_port, handler=accept)
  File "/Users/foouser/.virtualenvs/project/lib/python2.7/site-packages/paramiko/transport.py", line 810, in request_port_forward
    raise SSHException('TCP forwarding request denied')
paramiko.SSHException: TCP forwarding request denied

更新

根据@scott-talbert 的回答，我能够在设置 SSH 隧道后使用以下方法获得第一种工作方法。

from suds.client import Client
import os

url = 'http://{}/scripts/WebObjects.exe/WebServices.woa/ws/Law?wsdl'.format(os.getenv('SOAP_API_HOST'))
client = Client(url)
client.set_options(proxy={'http': '127.0.0.1:4000'})
print(client.service.doThing())

弄清楚如何让我的第二种方法发挥作用仍然很好，这样您就不必设置和管理 SSH 隧道。

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1 回答 1

1

我怀疑 #1 失败的原因是 WSDL 包含您的客户端无法直接访问的 URL，这就是您收到“连接被拒绝”消息的原因。看起来可以配置 Suds 告诉其 urllib2 实例使用代理服务器 - 请参阅https://fedorahosted.org/suds/wiki/Documentation。这可能适用于您的情况。您可能必须使用 -D 选项在“SOCKS”模式下运行 ssh。

于 2013-11-16T04:11:12.110 回答