How to send a xml-rpc request in python?

Question

I was just wondering, how would I be able to send a xml-rpc request in python? I know you can use xmlrpclib, but how do I send out a request in xml to access a function?

I would like to see the xml response.

So basically I would like to send the following as my request to the server:

<?xml version="1.0"?>
<methodCall>
  <methodName>print</methodName>
  <params>
    <param>
        <value><string>Hello World!</string></value>
    </param>
  </params>
</methodCall>

and get back the response

Eli Bendersky · Accepted Answer · 2010-02-11 04:06:34Z

13

Here's a simple XML-RPC client in Python:

import xmlrpclib

s = xmlrpclib.ServerProxy('http://localhost:8000')
print s.myfunction(2, 4)

Works with this server:

from SimpleXMLRPCServer import SimpleXMLRPCServer
from SimpleXMLRPCServer import SimpleXMLRPCRequestHandler

# Restrict to a particular path.
class RequestHandler(SimpleXMLRPCRequestHandler):
    rpc_paths = ('/RPC2',)

# Create server
server = SimpleXMLRPCServer(("localhost", 8000),
                            requestHandler=RequestHandler)

def myfunction(x, y):
    status = 1
    result = [5, 6, [4, 5]]
    return (status, result)
server.register_function(myfunction)

# Run the server's main loop
server.serve_forever()

To access the guts of xmlrpclib, i.e. looking at the raw XML requests and so on, look up the xmlrpclib.Transport class in the documentation.

answered Feb 11, 2010 at 4:06

Eli Bendersky

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3 Comments

John Jiang Over a year ago

Yes, I have that working, but how would i be able to send a request simply by using xml? And then getting the response back in xml?

Eli Bendersky Over a year ago

@PCBEEF: see the Transport class, as I said in the bottom. Examples are in the documentation - and you can also always look under the hood (the sources)

John Jiang Over a year ago

Found that setting the verbose=True flag when initiating the client, outputs the raw request and response

Nicholas Clarke · Accepted Answer · 2016-06-16 07:18:48Z

I have pared down the source code in xmlrpc.client to a minimum required to send a xml rpc request (as I was interested in trying to port the functionality). It returns the response XML.

Server:

from xmlrpc.server import SimpleXMLRPCServer

def is_even(n):
    return n%2 == 0

server = SimpleXMLRPCServer(("localhost", 8000))
print("Listening on port 8000...")
server.register_function(is_even, "is_even")
server.serve_forever()

Client:

import http.client

request_body = b"<?xml version='1.0'?>\n<methodCall>\n<methodName>is_even</methodName>\n<params>\n<param>\n<value><int>2</int></value>\n</param>\n</params>\n</methodCall>\n"

connection = http.client.HTTPConnection('localhost:8000')
connection.putrequest('POST', '/')
connection.putheader('Content-Type', 'text/xml')
connection.putheader('User-Agent', 'Python-xmlrpc/3.5')
connection.putheader("Content-Length", str(len(request_body)))
connection.endheaders(request_body)

print(connection.getresponse().read())

Alex Martelli · Accepted Answer · 2010-02-11 04:08:34Z

1

What do you mean by "get around"? xmlrpclib is the normal way to write an XML-RPC client in python. Just look at the sources (or copy them to your own module and add print statements!-) if you want to know the details of how things are done.

answered Feb 11, 2010 at 4:08

Alex Martelli

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