Python: Is It Possible To Set The Clientport With Xmlrpclib?
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 li
Solution 1:
Try to define a custom transport. This should be something like that:
import xmlrpclib, httplib
classsourcedTransport(xmlrpclib.Transport):defsetSource(self, src):
self.src = src
defmake_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)
Solution 2:
There is no option for this in module xmlrpclib, but you can create your own by modifying the original version. Assuming you use Linux, fetch /usr/lib/python2.7/xmlrpclib.py
. Modify the method make_connection
accordingly.
Providing a parameter source_address
to HTTPConnection
is supported by httplib not before Python version 2.7.
Have fun!
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