[Twisted-Python] How to send response 401

[Justin Johnson]

I have an xmlrpc service inheriting from xmlrpc.XMLRPC.  I override
render to allow for basic authentication as show below.  This all works,
except that I don't know what to call to send response 401 back to the
client.  self.send_response(401) is how I would do it with the
SimpleXMLRPCServer, but obviously doesn't work with the twisted xmlrpc
server.  Can someone point me in the right direction on this?

Thanks much.
-Justin

P.S. - Moshe, I figured out everything else I asked you on the
python-list about this.  Just need to know how to send response 401 now. 
Before I wasn't seeing the header, because request.headers doesn't
contain all headers.  Had to use request.getAllHeaders().


	def render(self, request):
		headers = request.getAllHeaders()
		if headers.has_key("authorization"):
			real_auth = '%s:%s' % (config.adminUser, config.adminPassword)
			auth = headers["authorization"]
			auth = auth.replace("Basic ","")
			decoded_auth = decodestring(auth)
			if decoded_auth == real_auth:
				return xmlrpc.XMLRPC.render(self, request)
		else:
			self.send_response(401)
			self.end_headers()