Python：是否可以使用 xmlrpclib 设置客户端端口？

发布于 2024-12-02 11:23:17 字数 249 浏览 2 评论 0原文

是否可以设置 xmlrpc 连接的客户端端口？

我想说：

客户端应该通过特定的客户端端口创建一个 ServerProxy 对象

或伪代码创建一个 ServerProxy 对象，如下所示：

serv = xmlrpclib.ServerProxy("server:port","overSpecificClientPort").

原文

Is it possible to set the clientport for the xmlrpc-connection?

I want to say:

Client should make a ServerProxy-object to over a specific client port

or pseudocode something like this:

serv = xmlrpclib.ServerProxy("server:port","overSpecificClientPort").

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绅刃 2024-12-09 11:23:17

尝试定义自定义传输。这应该是这样的：

import xmlrpclib, httplib

class sourcedTransport(xmlrpclib.Transport):
    def setSource(self, src):
        self.src = src
    def make_connection(self, host):
        h = httplib.HTTPConnection(host, source_address= self.src)
        return h

srcPort = 43040
srcAddress = ('', srcPort)
p = sourcedTransport()
p.setSource(srcAddress)
server = xmlrpclib.ServerProxy("server:port", transport=p)

编辑：错误修复 httplib.HTTP => httplib.HTTPConnection

并检查它是否有效，在 python 2.7 中（但不是之前）

Try to define a custom transport. This should be something like that:

import xmlrpclib, httplib

class sourcedTransport(xmlrpclib.Transport):
    def setSource(self, src):
        self.src = src
    def make_connection(self, host):
        h = httplib.HTTPConnection(host, source_address= self.src)
        return h

srcPort = 43040
srcAddress = ('', srcPort)
p = sourcedTransport()
p.setSource(srcAddress)
server = xmlrpclib.ServerProxy("server:port", transport=p)

EDIT: bug fix httplib.HTTP => httplib.HTTPConnection

And checked that it works, in python 2.7 (but not before)

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