12

我正在尝试与 交互supervisord，我想通过 unix 套接字（它是一个共享托管环境）与它交谈。

到目前为止我尝试过的是：

import xmlrpclib
server = xmlrpclib.ServerProxy('unix:///path/to/supervisor.sock/RPC2')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/usr/local/lib/python2.7/xmlrpclib.py", line 1549, in __init__
    raise IOError, "unsupported XML-RPC protocol"
IOError: unsupported XML-RPC protocol

/path/to/supervisor.sock肯定存在。'unix:///path/to/supervisor.sock/RPC2' 形式的 URI 由 使用supervisord，这就是我想到的地方。文档不讨论 unix 套接字：http ://docs.python.org/library/xmlrpclib.html 。

这可能吗？我应该使用不同的库吗？

4

4 回答 4

17

xmlrpclib要求传递的 url 以httpor开头https。解决这个问题的方法是定义一个忽略该 url 的自定义传输。以下是使用来自主管的传输的一些代码：

import supervisor.xmlrpc
import xmlrpclib

proxy = xmlrpclib.ServerProxy('http://127.0.0.1',
                               transport=supervisor.xmlrpc.SupervisorTransport(
                                    None, None, serverurl='unix://'+socketpath))

proxy.supervisor.getState()

如果这没有用，这里是代码的更新版本：

class UnixStreamHTTPConnection(httplib.HTTPConnection, object):
    def __init__(self, *args, **kwargs):
        self.socketpath = kwargs.pop('socketpath')
        super(UnixStreamHTTPConnection, self).__init__(*args, **kwargs)

    def connect(self):
        self.sock = socket.socket(socket.AF_UNIX, socket.SOCK_STREAM)
        self.sock.connect_ex(self.socketpath)

class UnixStreamTransport(xmlrpclib.Transport, object):
    def __init__(self, *args, **kwargs):
        self.socketpath = kwargs.pop('socketpath')
        super(UnixStreamTransport, self).__init__(*args, **kwargs)
于 2012-07-31T17:47:48.563 回答
8

我的python3版本，从上面的版本准备的。

from http.client import HTTPConnection
import socket
from xmlrpc import client

class UnixStreamHTTPConnection(HTTPConnection):
    def connect(self):
        self.sock = socket.socket(socket.AF_UNIX, socket.SOCK_STREAM)
        self.sock.connect(self.host)

class UnixStreamTransport(client.Transport, object):
    def __init__(self, socket_path):
        self.socket_path = socket_path
        super(UnixStreamTransport, self).__init__()

    def make_connection(self, host):
        return UnixStreamHTTPConnection(self.socket_path)

proxy = client.ServerProxy('http://localhost', transport=UnixStreamTransport("/var/run/supervisor.sock"))

print(proxy.supervisor.getState())
于 2018-07-17T09:00:01.623 回答
7

这是使用 xmlrpclib 与主管对话的更新示例：

import httplib
import socket
import xmlrpclib

class UnixStreamHTTPConnection(httplib.HTTPConnection):
    def connect(self):
        self.sock = socket.socket(socket.AF_UNIX, socket.SOCK_STREAM)
        self.sock.connect(self.host)

class UnixStreamTransport(xmlrpclib.Transport, object):
    def __init__(self, socket_path):
        self.socket_path = socket_path
        super(UnixStreamTransport, self).__init__()

    def make_connection(self, host):
        return UnixStreamHTTPConnection(self.socket_path)


server = xmlrpclib.Server('http://arg_unused', transport=UnixStreamTransport("/var/run/supervisor.sock"))
print(server.supervisor.getState())

正如已经提到的，我们必须使用 http:// 或 https:// 指定一个虚拟 url，然后指定一个自定义传输来处理域套接字

于 2014-05-23T19:32:45.093 回答
3

混合上面的答案，这对我有用......

import httplib
import socket
import xmlrpclib

class UnixStreamHTTPConnection(httplib.HTTPConnection, object):
    def __init__(self, *args, **kwargs):
        self.socketpath = kwargs.pop('socketpath')
        super(UnixStreamHTTPConnection, self).__init__(*args, **kwargs)

    def connect(self):
        self.sock = socket.socket(socket.AF_UNIX, socket.SOCK_STREAM)
        self.sock.connect(self.socketpath)

class UnixStreamTransport(xmlrpclib.Transport, object):
    def __init__(self, *args, **kwargs):
        self.socketpath = kwargs.pop('socketpath')
        super(UnixStreamTransport, self).__init__(*args, **kwargs)

    def make_connection(self, host):
        return UnixStreamHTTPConnection(host, socketpath=self.socketpath)

server = xmlrpclib.ServerProxy('http://arg_unused', transport=UnixStreamTransport(socketpath="path/to/supervisor.sock"))
print server.supervisor.getState()
于 2017-07-14T13:42:47.527 回答