I am intermittently receiving a httplib.CannotSendRequest exception when using a chain of SimpleXMLRPCServers that use the SocketServer.ThreadingMixin.
What I mean by 'chain' is the following:
I have a client script which uses xmlrpclib to call a function on a SimpleXMLRPCServer. That server, in turn, calls another SimpleXMLRPCServer. I realise how convoluted that sounds, but there are good reasons that this architecture has been selected, and I don't see a reason it shouldn't be possible.
(testclient)client_script ---calls-->
(middleserver)SimpleXMLRPCServer ---calls--->
(finalserver)SimpleXMLRPCServer --- does something
I have been able to reproduce the issue in the simple test code below. There are three snippets:
finalserver:
import SocketServer
import time
from SimpleXMLRPCServer import SimpleXMLRPCServer
from SimpleXMLRPCServer import SimpleXMLRPCRequestHandler
class AsyncXMLRPCServer(SocketServer.ThreadingMixIn,SimpleXMLRPCServer): pass
# Create server
server = AsyncXMLRPCServer(('', 9999), SimpleXMLRPCRequestHandler)
server.register_introspection_functions()
def waste_time():
time.sleep(10)
return True
server.register_function(waste_time, 'waste_time')
server.serve_forever()
middleserver:
import SocketServer
from SimpleXMLRPCServer import SimpleXMLRPCServer
from SimpleXMLRPCServer import SimpleXMLRPCRequestHandler
import xmlrpclib
class AsyncXMLRPCServer(SocketServer.ThreadingMixIn,SimpleXMLRPCServer): pass
# Create server
server = AsyncXMLRPCServer(('', 8888), SimpleXMLRPCRequestHandler)
server.register_introspection_functions()
s = xmlrpclib.ServerProxy('http://localhost:9999')
def call_waste():
s.waste_time()
return True
server.register_function(call_waste, 'call_waste')
server.serve_forever()
testclient:
import xmlrpclib
s = xmlrpclib.ServerProxy('http://localhost:8888')
print s.call_waste()
To reproduce, the following steps should be used:
Quite often (almost every time) you will get the error below the first time you try to run step 4. Interestingly, if you immediately try to run step (4) again the error will not occur.
Traceback (most recent call last):
File "testclient.py", line 6, in <module>
print s.call_waste()
File "/usr/lib64/python2.7/xmlrpclib.py", line 1224, in __call__
return self.__send(self.__name, args)
File "/usr/lib64/python2.7/xmlrpclib.py", line 1578, in __request
verbose=self.__verbose
File "/usr/lib64/python2.7/xmlrpclib.py", line 1264, in request
return self.single_request(host, handler, request_body, verbose)
File "/usr/lib64/python2.7/xmlrpclib.py", line 1297, in single_request
return self.parse_response(response)
File "/usr/lib64/python2.7/xmlrpclib.py", line 1473, in parse_response
return u.close()
File "/usr/lib64/python2.7/xmlrpclib.py", line 793, in close
raise Fault(**self._stack[0])
xmlrpclib.Fault: <Fault 1: "<class 'httplib.CannotSendRequest'>:">
The internet appears to say that this exception can be caused by multiple calls to httplib.HTTPConnection.request without intervening getresponse calls. However, the internet doesn't discuss this in the context of SimpleXMLRPCServer. Any pointers in the direction of resolving the httplib.CannotSendRequest issue would be appreciated.
=========================================================================================== ANSWER:
Okay, I'm a bit stupid. I think I was staring at the code for too protracted a period of time that I missed the obvious solution staring me in the face (quite literally, because the answer is actually in the actual question.)
Basically, the CannotSendRequest occurs when an httplib.HTTPConnection is interrupted by an intervening 'request' operation. Each httplib.HTTPConnection.request must be paired with a .getresponse() call. If that pairing is interrupted by another request operation, the second request will produce the CannotSendRequest error. so:
connection = httplib.HTTPConnection(...)
connection.request(...)
connection.request(...)
will fail because you have two requests on the same connection before any getresponse is called.
Linking that back to my question:
The solution then, is obviously to have each thread create it's own serverproxy. The fixed version of middleserver is below, and it works:
import SocketServer
from SimpleXMLRPCServer import SimpleXMLRPCServer
from SimpleXMLRPCServer import SimpleXMLRPCRequestHandler
import xmlrpclib
class AsyncXMLRPCServer(SocketServer.ThreadingMixIn,SimpleXMLRPCServer): pass
# Create server
server = AsyncXMLRPCServer(('', 8888), SimpleXMLRPCRequestHandler)
server.register_introspection_functions()
def call_waste():
# Each call to this function creates its own serverproxy.
# If this function is called by concurrent threads, each thread
# will safely have its own serverproxy.
s = xmlrpclib.ServerProxy('http://localhost:9999')
s.waste_time()
return True
server.register_function(call_waste, 'call_waste')
server.serve_forever()
Since this version results in each thread having its own xmlrpclib.serverproxy, there is no risk of the same instance of serverproxy invoking HTTPConnection.request more than once in succession. The programs work as intended.
Sorry for the bother.
Okay, I'm a bit stupid. I think I was staring at the code for to protracted a period of time that I missed the obvious solution staring me in the face (quite literally, because the answer is actually in the actual question.)
Basically, the CannotSendRequest occurs when an httplib.HTTPConnection is interrupted by an intervening 'request' operation. Basically, each httplib.HTTPConnection.request must be paired with a .getresponse() call. If that pairing is interrupted by another request operation, the second request will produce the CannotSendRequest error. so:
connection = httplib.HTTPConnection(...)
connection.request(...)
connection.request(...)
will fail because you have two requests on the same connection before any getresponse is called.
Linking that back to my question:
The solution then, is obviously to have each thread create it's own serverproxy. The fixed version of middleserver is below, and it works:
import SocketServer
from SimpleXMLRPCServer import SimpleXMLRPCServer
from SimpleXMLRPCServer import SimpleXMLRPCRequestHandler
import xmlrpclib
class AsyncXMLRPCServer(SocketServer.ThreadingMixIn,SimpleXMLRPCServer): pass
# Create server
server = AsyncXMLRPCServer(('', 8888), SimpleXMLRPCRequestHandler)
server.register_introspection_functions()
def call_waste():
# Each call to this function creates its own serverproxy.
# If this function is called by concurrent threads, each thread
# will safely have its own serverproxy.
s = xmlrpclib.ServerProxy('http://localhost:9999')
s.waste_time()
return True
server.register_function(call_waste, 'call_waste')
server.serve_forever()
Since this version results in each thread having its own xmlrpclib.serverproxy, there is no risk of the serverproxy invoking HTTPConnection.request more than once in succession. The programs work as intended.
Sorry for the bother.