Develop Reference

How to send multiple messages over same socket connection?

I am trying to send an array of messages through the same socket connection, but I get an error.
Here is my client code:
def send_over_socket(hl7_msg_array):
sock = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
sock.connect((config.HOST, config.PORT))
for single_hl7_msg in hl7_msg_array:
sock.send(single_hl7_msg.to_mllp().encode('UTF-8'))
received = sock.recv(1024*1024)
print("Sent: ", received)
sock.shutdown()
sock.close()
While debugging the code, I see that the exception occurs when I call the sock.recv(1024*1024) for the second message.
Here is the error:
ConnectionAbortedError: [WinError 10053] An established connection was aborted by the software in your host machine
Server-side code:
def run_mllp_server():
class PDQHandler(AbstractHandler):
def reply(self):
msg = hl7.parse(self.incoming_message)
msg_pid = msg[1][3]
msg_key = msg[2][3][0][1]
msg_value = msg[2][5]
lock = RLock()
lock.acquire()
results_collection[str(msg_pid)][str(msg_key)] = str(msg_value)
lock.release()
print("Received: ", repr(self.incoming_message))
return parse_message(self.incoming_message).to_mllp()
# error handler
class ErrorHandler(AbstractErrorHandler):
def reply(self):
if isinstance(self.exc, UnsupportedMessageType):
print("Error handler success 1")
else:
print("Error handler else case")
handlers = {
'ORU^R01^ORU_R01': (PDQHandler,),
'ERR': (ErrorHandler,)
}
server = MLLPServer(config.SOCKET_HOST, config.SOCKET_PORT, handlers)
print("Running Socket on port ", config.SOCKET_PORT)
server.serve_forever()
Here I am using MLLP protocol which has a TCP connection behind the scenes.
Can you help me please figure out what is going on? Is it a problem of ACK?

I do not know python at all but...
I do not think multiple messages is your problem. Looking at exception, I guess your first message is being sent correctly. Then, your client code waits for ACK to be received; but server never sends it. It instead closes the connection.
Also, make sure that whether sendall should be used instead of send.
After above issue is fixed, to send multiple messages on same connection, you have to follow MLLP (also called LLP) so that server can differentiate the message.
Description HEX ASCII Symbol
Message starting character 0B 11 <VT>
Message ending characters 1C,0D 28,13 <FS>,<CR>
This way, when you send a message to Listener (TCP/MLLP server), it looks for Start and End Block in your incoming data. Based on it, it differentiates each message.
Please refer to this answer for more details.

Python: Multithreaded socket server runs endlessly when client stops unexpectedly

I have created a multithreaded socket server to connect many clients to the server using python. If a client stops unexpectedly due to an exception, server runs nonstop. Is there a way to kill that particular thread alone in the server and the rest running
Server:
class ClientThread(Thread):
def __init__(self,ip,port):
Thread.__init__(self)
self.ip = ip
self.port = port
print("New server socket thread started for " + ip + ":" + str(port))
def run(self):
while True :
try:
message = conn.recv(2048)
dataInfo = message.decode('ascii')
print("recv:::::"+str(dataInfo)+"::")
except:
print("Unexpected error:", sys.exc_info()[0])
Thread._stop(self)
tcpServer = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
tcpServer.setsockopt(socket.SOL_SOCKET, socket.SO_REUSEADDR, 1)
tcpServer.bind((TCP_IP, 0))
tcpServer.listen(10)
print("Port:"+ str(tcpServer.getsockname()[1]))
threads = []
while True:
print( "Waiting for connections from clients..." )
(conn, (ip,port)) = tcpServer.accept()
newthread = ClientThread(ip,port)
newthread.start()
threads.append(newthread)
for t in threads:
t.join()
Client:
def Main():
s = socket.socket(socket.AF_INET,socket.SOCK_STREAM)
s.connect((host,int(port)))
while True:
try:
message = input("Enter Command")
s.send(message.encode('ascii'))
except Exception as ex:
logging.exception("Unexpected error:")
break
s.close()

Sorry about a very, very long answer but here goes.
There are quite a many issues with your code. First of all, your client does not actually close the socket, as s.close() will never get executed. Your loop is interrupted at break and anything that follows it will be ignored. So change the order of these statements for the sake of good programming but it has nothing to do with your problem.
Your server code is wrong in quite a many ways. As it is currently written, it never exits. Your threads also do not work right. I have fixed your code so that it is a working, multithreaded server, but it still does not exit as I have no idea what would be the trigger to make it exit. But let us start from the main loop:
while True:
print( "Waiting for connections from clients..." )
(conn, (ip,port)) = tcpServer.accept()
newthread = ClientThread(conn, ip,port)
newthread.daemon = True
newthread.start()
threads.append(newthread) # Do we need this?
for t in threads:
t.join()
I have added passing of conn to your client thread, the reason of which becomes apparent in a moment. However, your while True loop never breaks, so you will never enter the for loop where you join your threads. If your server is meant to be run indefinitely, this is not a problem at all. Just remove the for loop and this part is fine. You do not need to join threads just for the sake of joining them. Joining threads only allows your program to block until a thread has finished executing.
Another addition is newthread.daemon = True. This sets your threads to daemonic, which means they will exit as soon as your main thread exits. Now your server responds to control + c even when there are active connections.
If your server is meant to be never ending, there is also no need to store threads in your main loop to threads list. This list just keeps growing as a new entry will be added every time a client connects and disconnects, and this leaks memory as you are not using the threads list for anything. I have kept it as it was there, but there still is no mechanism to exit the infinite loop.
Then let us move on to your thread. If you want to simplify the code, you can replace the run part with a function. There is no need to subclass Thread in this case, but this works so I have kept your structure:
class ClientThread(Thread):
def __init__(self,conn, ip,port):
Thread.__init__(self)
self.ip = ip
self.port = port
self.conn = conn
print("New server socket thread started for " + ip + ":" + str(port))
def run(self):
while True :
try:
message = self.conn.recv(2048)
if not message:
print("closed")
try:
self.conn.close()
except:
pass
return
try:
dataInfo = message.decode('ascii')
print("recv:::::"+str(dataInfo)+"::")
except UnicodeDecodeError:
print("non-ascii data")
continue
except socket.error:
print("Unexpected error:", sys.exc_info()[0])
try:
self.conn.close()
except:
pass
return
First of all, we store conn to self.conn. Your version used a global version of conn variable. This caused unexpected results when you had more than one connection to the server. conn is actually a new socket created for the client connection at accept, and this is unique to each thread. This is how servers differentiate between client connections. They listen to a known port, but when the server accepts the connection, accept creates another port for that particular connection and returns it. This is why we need to pass this to the thread and then read from self.conn instead of global conn.
Your server "hung" upon client connetion errors as there was no mechanism to detect this in your loop. If the client closes connection, socket.recv() does not raise an exception but returns nothing. This is the condition you need to detect. I am fairly sure you do not even need try/except here but it does not hurt - but you need to add the exception you are expecting here. In this case catching everything with undeclared except is just wrong. You have also another statement there potentially raising exceptions. If your client sends something that cannot be decoded with ascii codec, you would get UnicodeDecodeError (try this without error handling here, telnet to your server port and copypaste some Hebrew or Japanese into the connection and see what happens). If you just caught everything and treated as socket errors, you would now enter the thread ending part of the code just because you could not parse a message. Typically we just ignore "illegal" messages and carry on. I have added this. If you want to shut down the connection upon receiving a "bad" message, just add self.conn.close() and return to this exception handler as well.
Then when you really are encountering a socket error - or the client has closed the connection, you will need to close the socket and exit the thread. You will call close() on the socket - encapsulating it in try/except as you do not really care if it fails for not being there anymore.
And when you want to exit your thread, you just return from your run() loop. When you do this, your thread exits orderly. As simple as that.
Then there is yet another potential problem, if you are not only printing the messages but are parsing them and doing something with the data you receive. This I do not fix but leave this to you.
TCP sockets transmit data, not messages. When you build a communication protocol, you must not assume that when your recv returns, it will return a single message. When your recv() returns something, it can mean one of five things:
The client has closed the connection and nothing is returned
There is exactly one full message and you receive that
There is only a partial message. Either because you read the socket before the client had transmitted all data, or because the client sent more than 2048 bytes (even if your client never sends over 2048 bytes, a malicious client would definitely try this)
There are more than one messages waiting and you received them all
As 4, but the last message is partial.
Most socket programming mistakes are related to this. The programmer expects 2 to happen (as you do now) but they do not cater for 3-5. You should instead analyse what was received and act accordingly. If there seems to be less data than a full message, store it somewhere and wait for more data to appear. When more data appears, concatenate these and see if you now have a full message. And when you have parsed a full message from this buffer, inspect the buffer to see if there is more data there - the first part of the next message or even more full messages if your client is fast and server is slow. If you process a message and then wipe the buffer, you might have wiped also bytes from your next message.

Why doesn't my code respond elsewise?

import socket
sock = socket.socket()
sock.bind(('127.0.0.1', 1600))
sock.listen(1)
(client_sock,client_address) = sock.accept()
client_name = client_sock.recv(1024)
client_sock.send('hi its '+client_name)
client_sock.close()
sock.close();
So i run on cmd command prompt the file containing the code above, and all i get is this:
As it can be seen my cmd is as if waiting for something to happen, it doesnt print out "Hi its ..." and never ends from waiting.
Why doesn't my code work ?

It is simple: you have first to send the data and then receive. The receive command wait until the timeout (if it is set) and then go over. Remember that you have also to print the received data.
The best practice for doing what you want to do is a server-client approach having at least two different threads.

ZeroMQ push socket causes client to not terminate when no process is listening

I'm just starting to mess with ZeroMQ and I have a problem with a client that doesn't terminate normally. In particular I have a client that may "push" data when no sink server is listening and that seems to make the process hang after the python code has finished. I assume there is some background thread that needs to be cleaned up -- please tell me how or point to documentation.
Here is the relevant piece of code. If I run the process with no listener with the "self.push" line uncommented the process hangs
def setup(self):
print self.name, "connect to sockets"
ctx = self.ctx = zmq.Context()
self.pull = ctx.socket(zmq.PULL)
self.pull.connect(self.ventillatorAddress)
self.push = ctx.socket(zmq.PUSH)
self.push.connect(self.sinkAddress)
self.control = ctx.socket(zmq.SUB)
self.control.connect(self.publisherAddress)
self.control.setsockopt(zmq.SUBSCRIBE, "") # get every control message
self.inbox = ctx.socket(zmq.SUB)
self.inbox.connect(self.distributorAddress)
self.inbox.setsockopt(zmq.SUBSCRIBE, self.name) # listen only for messages addressed with name
def start(self):
print self.name, "push worker is ready signal"
# listen for "go" signal
pollcount = 0
go = False
while not go:
#print "send ready for", self.name
#self.push.send(self.name+" ready")
print "listen for 'go'"
msg = self.recvPoll(self.control)
if msg is None:
pollcount += 1
assert pollcount<10
print "poll timeout", pollcount
time.sleep(1)
continue
pollcount = 0
print "recv'd", msg
assert msg=="go!"
go = True
print "go signal received"
pass
With the line commented (and no listener) the process completes normally.
I tried context.term() and context.destroy() and they don't seem to help.
How can I clean up the socket? Or any other clues? Thanks in advance!

This is most probably due to the linger functionality of ZeroMQ. Quoting from the man page:
The ZMQ_LINGER option shall set the linger period for the specified socket. The linger period determines how long pending messages which have yet to be sent to a peer shall linger in memory after a socket is closed with zmq_close(3), and further affects the termination of the socket's context with zmq_term(3).
The default value causes ZeroMQ to wait indefinitely until it is able to deliver the stuck message.
Try setting the ZMQ_LINGER socket option to zero or a short time (in milliseconds).

zeromq: how to prevent infinite wait?

I just got started with ZMQ. I am designing an app whose workflow is:
one of many clients (who have random PULL addresses) PUSH a request to a server at 5555
the server is forever waiting for client PUSHes. When one comes, a worker process is spawned for that particular request. Yes, worker processes can exist concurrently.
When that process completes it's task, it PUSHes the result to the client.
I assume that the PUSH/PULL architecture is suited for this. Please correct me on this.
But how do I handle these scenarios?
the client_receiver.recv() will wait for an infinite time when server fails to respond.
the client may send request, but it will fail immediately after, hence a worker process will remain stuck at server_sender.send() forever.
So how do I setup something like a timeout in the PUSH/PULL model?
EDIT: Thanks user938949's suggestions, I got a working answer and I am sharing it for posterity.

If you are using zeromq >= 3.0, then you can set the RCVTIMEO socket option:
client_receiver.RCVTIMEO = 1000 # in milliseconds
But in general, you can use pollers:
poller = zmq.Poller()
poller.register(client_receiver, zmq.POLLIN) # POLLIN for recv, POLLOUT for send
And poller.poll() takes a timeout:
evts = poller.poll(1000) # wait *up to* one second for a message to arrive.
evts will be an empty list if there is nothing to receive.
You can poll with zmq.POLLOUT, to check if a send will succeed.
Or, to handle the case of a peer that might have failed, a:
worker.send(msg, zmq.NOBLOCK)
might suffice, which will always return immediately - raising a ZMQError(zmq.EAGAIN) if the send could not complete.

This was a quick hack I made after I referred user938949's answer and http://taotetek.wordpress.com/2011/02/02/python-multiprocessing-with-zeromq/ . If you do better, please post your answer, I will recommend your answer.
For those wanting lasting solutions on reliability, refer http://zguide.zeromq.org/page:all#toc64
Version 3.0 of zeromq (beta ATM) supports timeout in ZMQ_RCVTIMEO and ZMQ_SNDTIMEO. http://api.zeromq.org/3-0:zmq-setsockopt
Server
The zmq.NOBLOCK ensures that when a client does not exist, the send() does not block.
import time
import zmq
context = zmq.Context()
ventilator_send = context.socket(zmq.PUSH)
ventilator_send.bind("tcp://127.0.0.1:5557")
i=0
while True:
i=i+1
time.sleep(0.5)
print ">>sending message ",i
try:
ventilator_send.send(repr(i),zmq.NOBLOCK)
print " succeed"
except:
print " failed"
Client
The poller object can listen in on many recieving sockets (see the "Python Multiprocessing with ZeroMQ" linked above. I linked it only on work_receiver. In the infinite loop, the client polls with an interval of 1000ms. The socks object returns empty if no message has been recieved in that time.
import time
import zmq
context = zmq.Context()
work_receiver = context.socket(zmq.PULL)
work_receiver.connect("tcp://127.0.0.1:5557")
poller = zmq.Poller()
poller.register(work_receiver, zmq.POLLIN)
# Loop and accept messages from both channels, acting accordingly
while True:
socks = dict(poller.poll(1000))
if socks:
if socks.get(work_receiver) == zmq.POLLIN:
print "got message ",work_receiver.recv(zmq.NOBLOCK)
else:
print "error: message timeout"

The send wont block if you use ZMQ_NOBLOCK, but if you try closing the socket and context, this step would block the program from exiting..
The reason is that the socket waits for any peer so that the outgoing messages are ensured to get queued.. To close the socket immediately and flush the outgoing messages from the buffer, use ZMQ_LINGER and set it to 0..

If you're only waiting for one socket, rather than create a Poller, you can do this:
if work_receiver.poll(1000, zmq.POLLIN):
print "got message ",work_receiver.recv(zmq.NOBLOCK)
else:
print "error: message timeout"
You can use this if your timeout changes depending on the situation, instead of setting work_receiver.RCVTIMEO.