Hello everyone!
I am trying to set up a broker-worker system (one broker, multiple and
heterogeneous workers) in python for a local network (though possibly
also over inet). I am using SimpleXMLRPCSer ver and
xmlrpclib.Serve rProxy for connections.
Whatever I do, after about 100,000 to 150,000 XMLRPCalls over a single
ServerProxy connection, I am getting a connection error (usually
"connection reset by peer"). I have tried deleting the ServerProxy,
setting up a new one and calling the function again when the error
occurs, but this does not change anything:
# this is the worker's secureRPC; the broker's works analogously
def worker_secureRP C(self, fname, args=None):
tries = 2 # raising the nr of tries does not change
anything
retval = None
for i in range(0,tries):
try:
fref = self.getfref(fn ame)
# returns a reference to
self.sender.fna me
if (args==None):
retval = fref()
else:
retval = fref(*args)
except:
self.sender = None
time.sleep(1)
self.sender = ServerProxy("ht tp://" + self.brokerHost + ":" +
str(self.broker Port))
if (i>=(tries-1)):
raise
Then, thinking that maybe the "return" statement produces the error, I
tried doing error handling there as well (keep in mind that the
function might have been called already, so don't do the same action
twice):
def worker_foo(self , popcontainer, failed=[]):
self.lock.acqui re()
if (len(failed) 0):
while (len(failed) 0):
failed.pop()
else:
do_stuff_in_new _thread()
self.lock.relea se()
try:
return True
except:
failed.append(" failed")
The function foo gets called remotely after the set-up of the new
ServerProxy, but does produce the same error again! (socket.error:
connection reset by peer). Btw, another function sends the result back
to the Broker again via xmlrpc, resulting in asynchronous
communication, so we only need to return some bogus value here.
Am I missing sth obvious here, or why can I not re-build the
connection? What can I do to handle this error correctly?
cheers,
Christoph
I am trying to set up a broker-worker system (one broker, multiple and
heterogeneous workers) in python for a local network (though possibly
also over inet). I am using SimpleXMLRPCSer ver and
xmlrpclib.Serve rProxy for connections.
Whatever I do, after about 100,000 to 150,000 XMLRPCalls over a single
ServerProxy connection, I am getting a connection error (usually
"connection reset by peer"). I have tried deleting the ServerProxy,
setting up a new one and calling the function again when the error
occurs, but this does not change anything:
# this is the worker's secureRPC; the broker's works analogously
def worker_secureRP C(self, fname, args=None):
tries = 2 # raising the nr of tries does not change
anything
retval = None
for i in range(0,tries):
try:
fref = self.getfref(fn ame)
# returns a reference to
self.sender.fna me
if (args==None):
retval = fref()
else:
retval = fref(*args)
except:
self.sender = None
time.sleep(1)
self.sender = ServerProxy("ht tp://" + self.brokerHost + ":" +
str(self.broker Port))
if (i>=(tries-1)):
raise
Then, thinking that maybe the "return" statement produces the error, I
tried doing error handling there as well (keep in mind that the
function might have been called already, so don't do the same action
twice):
def worker_foo(self , popcontainer, failed=[]):
self.lock.acqui re()
if (len(failed) 0):
while (len(failed) 0):
failed.pop()
else:
do_stuff_in_new _thread()
self.lock.relea se()
try:
return True
except:
failed.append(" failed")
The function foo gets called remotely after the set-up of the new
ServerProxy, but does produce the same error again! (socket.error:
connection reset by peer). Btw, another function sends the result back
to the Broker again via xmlrpc, resulting in asynchronous
communication, so we only need to return some bogus value here.
Am I missing sth obvious here, or why can I not re-build the
connection? What can I do to handle this error correctly?
cheers,
Christoph