I want to know in which manner zeroMQ read form multiple connections ?
For example:
If I have a server which is connected to multiple clients and receiving data at the same time from all the clients which of the one it will read first?
Is it round robin or some other algorithm?
Use pub-sub routine, Python example:
#Publishing script
import zmq
ctx = zmq.Context()
socket_publish = ctx.socket(zmq.PUB)
socket_publish.bind("tcp://*:7787") #define socket for publishing
#subscribing script(s)
ctx = zmq.Context()
s = ctx.socket(zmq.SUB)
s.connect("tcp://127.0.0.1:7787") #connect to the socket multiple times
s.setsockopt(zmq.SUBSCRIBE,'')
msg = s.recv()
Related
I'm sending video as a sequence of images (equals zmq messages) but sometimes, perhaps when the network is slow, they are received at a slower rate than they're sent and a growing latency appears, seemingly up to about a minute of video or 100s of images or megabytes of data. It usually clears itself eventually with the subscriber receiving messages at a faster rate than the publisher sends.
Instead, I want it to discard missed messages the same way it's supposed to if the subscriber is too slow recving them. I hoped zmq.CONFLATE=1 would do this but it doesn't. How then? I suspect they're being buffered at the publisher, which is not supposed to have any zmq buffer, or in the network stack somehow.
Simplified server code
context = zmq.Context()
socket = context.socket(zmq.PUB)
socket.bind("tcp://*:12345")
camera = PiCamera()
stream = io.BytesIO()
for _ in camera.capture_continuous(stream, 'jpeg', use_video_port=True):
stream.truncate()
stream.seek(0)
socket.send(stream.read())
stream.seek(0)
Simplified client code
# Initialization
self.context = zmq.Context()
self.video_socket = self.context.socket(zmq.SUB)
self.video_socket.setsockopt(zmq.CONFLATE, 1)
self.video_socket.setsockopt(zmq.SUBSCRIBE, b"")
self.video_socket.connect("tcp://" + ip_address + ":12345")
def get_image(self):
# Receive the latest image
poll_result = self.video_socket.poll(timeout=0)
if poll_result == zmq.POLLIN:
return self.video_socket.recv()
else:
return None
The publisher is on a Raspberry Pi and the subscriber is on Windows.
I am not sure which version of python zmq you are using but based on the underlying c++ libzmq you need to:
Set the ZMQ_SNDHWM socket option on the server socket
Set the ZMQ_RCVHWM socket option on the client socket.
These options limit the number of messages to queue per completed connection in the case of pub/sub. If the queue grows larger than the HWM (high water mark) the messages will be discarded.
Also turn off conflate as that will interfere with these options.
Also set zmq.CONFLATE=1 on the server to keep only the latest message in the send queue.
Before binding the server socket
socket.setsockopt(zmq.CONFLATE, 1)
For some reason I mistakenly thought the PUB socket didn't have a send queue but it does.
i am learning how to use grpc streams to exchange messages between clients and server in python. I found a base example that enables the simple message sending between server and client. I am trying to modify it so that i could keep track of all the clients connected to the grpc server (on the server side) and could do two things: 1) broadcast from server to all clients, 2) send message to a particular connected client.
Here is the .proto file
syntax = 'proto3';
service Scenario {
rpc Chat(stream DPong) returns (stream DPong) {}
}
message DPong {
string name = 1;
}
And here is the client.py that creates a daemon process to listen for incoming messages and waits for stdin for any outgoing messages
import threading
import grpc
import time
import scenario_pb2_grpc, scenario_pb2
# new changes
msgQueue = queue.Queue()
def run():
channel = grpc.insecure_channel('localhost:50052')
stub = scenario_pb2_grpc.ScenarioStub(channel)
print('client connected')
global queue
def inputStream():
while 1:
msg = input('>>Enter message\n>>')
yield scenario_pb2.DPong(name=msg)
input_stream = stub.Chat(inputStream())
def read_incoming():
while 1:
print('receivedFromServer: {}\n>>'.format(next(input_stream).name))
thread = threading.Thread(target=read_incoming)
thread.daemon = True
thread.start()
while 1:
time.sleep(1)
if __name__ == '__main__':
print('client starting ...')
run()
Below is the server.py
import random
import string
import threading
import grpc
import scenario_pb2_grpc
import scenario_pb2
import time
from concurrent import futures
clientList = []
class Scenario(scenario_pb2_grpc.ScenarioServicer):
def Chat(self, request_iterator, context):
clients = []
def stream():
while 1:
time.sleep(1)
msg = input('>>Enter message\n>>')
for i in clientList:
yield msg
output_stream = stream()
def read_incoming():
while 1:
received = next(request_iterator).name
if (context,request_iterator) not in clientList:
clientList.append((context, request_iterator))
print('receivedFromClient: {}'.format(received), len(clientList))
thread = threading.Thread(target=read_incoming)
thread.daemon = True
thread.start()
while 1:
msg = output_stream
yield scenario_pb2.DPong(name=next(msg))
if __name__ == '__main__':
server = grpc.server(futures.ThreadPoolExecutor(max_workers=10))
scenario_pb2_grpc.add_ScenarioServicer_to_server(
Scenario(), server)
server.add_insecure_port('[::]:50052')
server.start()
print('listening ...')
while 1:
time.sleep(1)
So far, i have tried to maintain a list object clientList that contains the context & request_iterator object of the client, and is updated every time a new client joins the server. But how do i set these object from the clientList before sending out an outgoing message? I have tried to iterate the list but the server sends the message to the same client (the last client heard from) a number of times instead of sending it to all the clients once.
Any help is highly appreciated!
This is certainly possible. The problem that you're running into here is that each call to Scenario.Chat on the server side corresponds to a single client connection. That is, this function is called when the streaming RPC starts and as soon as the function exits, the RPC ends.
So if you want n connected clients, you'll need n instances of Scenario.Chat running concurrently, each on its own thread. This does mean that the number of concurrently connected clients is limited by the size of the threadpool with which you instantiate your server.
So, let's say you have n threads in your server process dedicated to maintaining client connections. Then you need another n+1th thread (perhaps the main thread) determining when the server will broadcast a message to all clients (maybe by looking for input from STDIN?). When this extra thread determines that a message should be broadcast, it needs to communicate this intent to all of the threads maintaining connections to a client. There are many ways to make this happen. A threading.Condition and a global collections.deque, or a collections.deque per client connection (somewhat like channels between goroutines) would be two ways. The tricky bit here is ensuring that each client connection will receive the message regardless of how long the client connection thread takes to wake up and how many messages the n+1th thread decides to send in the interim.
If this is still unclear, I can follow up with some actual code demonstrating the idea.
You can spin up multiple ports in one application.
gRPC can be running in port 50011 and flask with socket.io can be running in port 8080
with python, you can use the flask framework and flask_socketio library in your server.py
eg server.py
from flask import Flask
from flask_socketio import SocketIO, emit
app = Flask(__name__)
socketio = SocketIO(app)
#app.route('/')
def index():
return "Hello, World!"
if __name__ == '__main__':
app.run(port=8080)
app.run(debug=True)
socketio.run(app)
instead of using gRPC streaming API, use WebSocket to broadcast to all connected clients and specific/selected clients using rooms.
eg
#socketio.on('message')
def handle_message(data):
// logic to send large data in chunks the logic should call the
// emit function in socket.io and emit an event that send the large
// data in chunks eg emit('my response', chunkData)
gRPC is primarily built for one client request and response and WebSocket is for multiple clients.
By using 0mq, I am trying to detect if I have made a successful connection to a PULL port, and if I can PUSH. However, it didn't work as I had expected, see the example code below. Poller will return immediately even remote peer hasn't been started to accept connections. Is there a way to fix it?
import sys
import zmq
context = zmq.Context()
pusher = context.socket(zmq.PUSH)
pusher.connect("tcp://localhost:5555")
poller = zmq.Poller()
poller.register(pusher, zmq.POLLOUT)
socks = dict(poller.poll(timeout=1000))
if pusher in socks and socks[pusher] == zmq.POLLOUT:
print("Pusher can push")
else:
print("Failed to connect, exit.")
sys.exit(1)
You would be allowed to send as long as you haven't reached the High Water Mark ( HWM ) of the sending socket - the number of messages allowed to pile up on the sender side.
By default it is set to 1000 as far as I remember.
/Søren
I want to use zeromq for several clients labeled as 1...n to pull receive from one same socket in server pushing messages containing one field as client id. When server push message labeled 1, only client 1 receives it.
one way is to generate the same number of server sockets for clients to one by one connect, which I think there may be a better solution. Thanks a lot.
Use "Publish/Subscribe" pattern and set socket options for filtering messages.
Code for SUB side:
context = zmq.Context()
socket = context.socket(zmq.SUB);
socket.connect ("tcp://localhost:%s" % port)
topicfilter = "10001"
socket.setsockopt(zmq.SUBSCRIBE, topicfilter)
string = socket.recv()
Code for PUB side:
context = zmq.Context()
socket = context.socket(zmq.PUB)
socket.bind("tcp://*:%s" % port)
topic = 10001
messagedata = random.randrange(1,215) - 80
socket.send("%d %d" % (topic, messagedata))
Examples and pattern describe here
Update
Another pattern which you may use Router-Req.
Common idea:
Examples for python here
I have read through the zguide but haven't found the kind of pattern I'm looking for:
There is one central server (with known endpoint) and many clients (which may come and go).
Clients keep sending hearbeats to the server, but they don't want the server to reply.
Server receives heartbeats, but it does not reply to clients.
Hearbeats sent when clients and server are disconnected should somehow be dropped to prevent a heartbeat flood when they go back online.
The closet I can think of is the DEALER-ROUTER pattern, but since this is meant to be used as an async REQ-REP pattern (no?), I'm not sure what would happen if the server just keep silent on incoming "requests." Also, the DEALER socket would block rather then start dropping heartbeats when the send High Water Mark is reached, which would still result in a heartbeat flood.
The PUSH/PULL pattern should give you what you need.
# Client example
import zmq
class Client(object):
def __init__(self, client_id):
self.client_id = client_id
ctx = zmq.Context.instance()
self.socket = ctx.socket(zmq.PUSH)
self.socket.connect("tcp://localhost:12345")
def send_heartbeat(self):
self.socket.send(str(self.client_id))
# Server example
import zmq
class Server(object):
def __init__(self):
ctx = zmq.Context.instance()
self.socket = ctx.socket(zmq.PULL)
self.socket.bind("tcp://*:12345") # close quote
def receive_heartbeat(self):
return self.socket.recv() # returns the client_id of the message's sender
This PUSH/PULL pattern works with multiple clients as you wish. The server should keep an administration of the received messages (i.e. a dictionary like {client_id : last_received} which is updated with datetime.utcnow() on each received message. And implement some housekeeping function to periodically check the administration for clients with old timestamps.