Re: XML RPC Problem....

[Fredrik Lundh]

Usman Ajmal wrote:

And i also fount that a typical system.auth call will look like:
>
POST /xmlrpc/clarens_server. py HTTP/1.0
Host: localhost
User-Agent: xmlrpclib.py/0.9.9 (by www.pythonware.com <http://www.pythonware. com>)
>
Content-Type: text/xml
Content-Length: 105
AUTHORIZATION: Basic MkhVTm9VazYxbXA rVEZLS0dCY2tIRl A3bjVzPQo6RnJvb SBi
<?xml version='1.0'?>
<methodCall>
<methodName>sys tem.auth</methodName>
>
<params>
</params>
</methodCall>
>
>
Problem is that i don't know how do i generate above xml system.auth
call. Can anyone please tell me how do call a function, setting the
header of the call too?

you need to plugin a custom transport. see this page for an example:

http://www.python.org/doc/lib/xmlrpc-client-example.html

in your case, it should be sufficient to override send_request, e.g.
(untested):

class SecureTransport (xmlrpclib.Tran sport):

def set_authorizati on(self, ustring, text_ucert):
self.authoriati on = encodestring(
"%s:%s" % (ustring,text_u cert)
)

def send_request(se lf, connection, handler, request_body):
connection.putr equest("POST", handler)
connection.puth eader("Authoriz ation",
"Basic %s" % self.authorizat ion
)

and instantiate the transport by doing

t = SecureTransport ()
t.set_authoriza tion(ustring, text_ucert)

before passing to the server proxy.

</F>