CODEWITHSUNDEEP
pythonmultithreadingprocessmultiprocessingrpc
Nathan
Nathan

Reputation: 360

How to have one process be a RPC client and server at the same time in python?

I'm building a p2p network and each process represents a peer. I'm using RPC for peer-to-peer communication. Each process (peer) needs to be able to receive messages from other peers and also send out messages. I already have a pretty simple server and client toy projects running.

server.py:

cache = []
def updateCache(elem):
    cache.append(elem)
    print(cache)
    return "message updated"

server = SimpleXMLRPCServer(("localhost", 6789))
server.register_function(updateCache, "updateCache")
server.serve_forever()

client.py:

proxy.updateCache(f'message is: {sys.argv[1]}')

My question is, how do I combine the two so that one process can do both? What would be the best way to do this in python? Thank you!

Upvotes: 0

Views: 560

Answers (1)

Booboo
Booboo

Reputation: 44108

Per the comment of @furas:

def server():
    from xmlrpc.server import SimpleXMLRPCServer

    cache = []
    def updateCache(elem):
        cache.append(elem)
        print(cache)
        return "message updated"

    with SimpleXMLRPCServer(('localhost', 6789)) as server:
        server.register_function(updateCache, "updateCache")
        server.serve_forever()

def client(msg):
    from xmlrpc.client import ServerProxy

    proxy = ServerProxy('http://localhost:6789')
    print(proxy.updateCache(f'message is: {msg}'))

if __name__ == '__main__':
    from threading import Thread
    import sys

    # The server will end as soon as the main thread ends:
    Thread(target=server, daemon=True).start()

    client(sys.argv[1])

python test.py Booboo

Prints:

['message is: Booboo']
127.0.0.1 - - [16/Oct/2022 09:03:53] "POST /RPC2 HTTP/1.1" 200 -
message updated

Upvotes: 1

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