Running example code does not connect to server about python-actioncable-zwei HOT 8 CLOSED

Hi.

I've tested exactly the code you provided and it worked without problems on my machine.
Have you tested to connect to the server with another client to see if the problem is really caused by ActionCableZwei?

Best

from python-actioncable-zwei.

Hi Tobias, thanks again. Connecting works with websocket-client and the code below. Meaning, the server shows it received a connect attempt. I'm stumped. It's the same URL. And I've allowed all remote origins in my rails server.

Debug Output when I connect using example code (I added one to the runforever area)

python3 ./my_actioncable.py
04/03/2017 09:53:14 PM my_actioncable INFO: Started
04/03/2017 09:53:14 PM monitor DEBUG: ACTIONCABLE MONITOR : Started.
04/03/2017 09:53:14 PM connection DEBUG: ACTIONCABLE CONNECTION : RUN FOREVER...
04/03/2017 09:53:14 PM connection DEBUG: ACTIONCABLE CONNECTION : Connect...
$ (end)

Here is the example code from websocket-client

import websocket
import thread
import time
import json

def on_message(ws, message):
    print(message)

def on_error(ws, error):
    print(error)

def on_close(ws):
    print("### closed ###")

def on_open(ws):
    msg = {'identifier': 'status_channel', 'command': 'message','message': 'FROM PYTHON'}
    

    
    def run(*args):
        for i in range(3):
            time.sleep(1)

            ws.send("{\"command\":\"subscribe\",\"identifier\":{\"channel\":\"MeshRelayChannel\"}}")
        time.sleep(1)
        ws.close()
        print("thread terminating...")
    thread.start_new_thread(run, ())


if __name__ == "__main__":
    websocket.enableTrace(True)
    ws = websocket.WebSocketApp("ws://localhost:3000/cable/",
                              on_message = on_message,
                              on_error = on_error,
                              on_close = on_close)
    ws.on_open = on_open
    ws.run_forever()

from python-actioncable-zwei.

Update: It works if I change line 54 of connection.py to call ws.run_forever() , instead of the internal function you have defined (_run_forever)

The thread isn't executing the ws instance, so nothing gets kicked off. But I'm not sure why this works for you and not me?

self.ws_thread = threading.Thread(name="APIConnectionThread_{}".format(uuid.uuid1()), target=self.ws.run_forever())

from python-actioncable-zwei.

You actually call the run_forever method in your code. So the websocket connection will now block your main thread. You have to pass the reference to the new thread, so without parentheses.

BTW The private method I use as thread loop does actually also nothing other than calling the run_forever method of the ws. See here.

Are you trying to call this program directly from the command line like in the example above?

from python-actioncable-zwei.

I see that now, your private method does end up calling the run_forever method of the ws. And yes, it does block the main thread (undesired!) but my server does receive responses.

Yes, I am trying to call this program directly. I am copying and pasting your code directly into a file, and then running 'python3 my_actioncable.py' Should I be running it differently, and I'm not understanding how to use this example?

I am running: python 3.6.1, macOS 10.12.3

from python-actioncable-zwei.

Then add this code at the end of your file:

while True:
  pass

The connection thread is actually a daemonic thread so it's going to be killed if the main thread finishes. If you call an infinite loop in the main thread this will actually never stop and your connection should be established.

from python-actioncable-zwei.

Ahh, I see, that makes total sense. Threading 101. Thanks so much for putting with a newbie python guy. I really appreciate that you wrote this.

from python-actioncable-zwei.

Your welcome :) Have fun!

Best

from python-actioncable-zwei.