python-websocket and socket.io namespace

Question

I would write a websocket client in python to connect to a server written with socket.io. My current code, taken from 1 is the following:

import websocket, httplib, sys, asyncore
def connect(server, port):

    print("connecting to: %s:%d" %(server, port))

    conn  = httplib.HTTPConnection(server + ":" + str(port))
    conn.request('POST','/socket.io/1/')
    resp  = conn.getresponse() 
    hskey = resp.read().split(':')[0]
    ws = websocket.WebSocket(
                'ws://'+server+':'+str(port)+'/socket.io/1/websocket/'+hskey,
                onopen   = _onopen,
                onmessage = _onmessage,
                onclose = _onclose)

    return ws

def _onopen():
    print("opened!")

def _onmessage(msg):
    print("msg: " + str(msg))

def _onclose():
    print("closed!")


if __name__ == '__main__':

    server = 'localhost'
    port = 8081

    ws = connect(server, port)

    try:
        asyncore.loop()
    except KeyboardInterrupt:
        ws.close()

My question is how do I connect to a specific namespace?

Thanks

Roy Hyunjin Han · Accepted Answer · 2012-08-10 23:14:08Z

2

You can use socketIO-client which is available on PyPI under the MIT license. It has support for different namespaces of a single socket.

from socketIO_client import SocketIO, BaseNamespace

class MainNamespace(BaseNamespace):

    def on_aaa(self, *args):
        print 'aaa', args

class ChatNamespace(BaseNamespace):

    def on_bbb(self, *args):
        print 'bbb', args

class NewsNamespace(BaseNamespace):

    def on_ccc(self, *args):
        print 'ccc', args

mainSocket = SocketIO('localhost', 8000, MainNamespace)
chatSocket = mainSocket.connect('/chat', ChatNamespace)
newsSocket = mainSocket.connect('/news', NewsNamespace)
mainSocket.wait()

answered Aug 10, 2012 at 23:14

Roy Hyunjin Han

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1 Comment

simonmorley Over a year ago

For those of you using this, I believe the format's change slightly recently. Read the docs if you're getting stuck. pypi.python.org/pypi/socketIO-client

Anderson · Accepted Answer · 2014-11-18 08:35:32Z

1

You can change the namespace more easily.

from socketIO_client import SocketIO, BaseNamespace
socket = SocketIO('192.168.4.47', 7777)
chat = socket.define(BaseNamespace, '/openchat')
chat.emit('echo', 'hello openchat my name is Anderson')

I've solved this problem in https://pypi.python.org/pypi/socketIO-client

hope you save your time a lot

answered Nov 18, 2014 at 8:35

community wiki

Anderson

Comments

abourget · Accepted Answer · 2012-06-23 21:05:10Z

0

I'd suggest you use Wireshark to sniff a connection made with Socket.IO and that you send the correct packets through your websocket connection to fake being a Socket.IO client.. and then implement the packets and messaging protocol of the Socket.IO layer.

There is basic documentation of the packet types in here:

http://gevent-socketio.readthedocs.org/en/latest/packet.html#module-socketio.packet

Also, you can read the test suite that shows the wire-level protocol (that you will probably need to implement):

https://github.com/abourget/gevent-socketio/blob/master/tests/test_packet.py

A specific namespace is specified in the packet of different types as you will see in the docs.

Hope this helps.

answered Jun 23, 2012 at 21:05

abourget

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