Django websockets implementation

Question

I'm new to python and Django and i'm trying to implement websockets in Django.
What i do is i'm following the steps described in websockets documentation
The problem is that the server side command described has to be run in console. When i run it from console it works, but i want to run it inside the Django view asynchronously with a GET request. When i try it the server raises an exception something like this RuntimeError: There is no current event loop in thread 'Thread-2'.
To be more specific I want to use the technology to show a real time logs. For example an oracle procedure performs an insert and server pushes it to a page with websockets.
Am I on a wrong path for implementing the described or can anyone suggest a right/better solution?

I'm on django version 1.9 implemented on both Django's development server and Uwsgi and Nginx server, python version 3.5.2 on RedHatEnterpriseServer Release: 6.7

UPDATE
The exact code from the above URL and i put it in the view.

def ws(request):

    async def time(websocket, path):
        while True:
            now = datetime.datetime.utcnow().isoformat() + 'Z'
            await websocket.send(now)
            await asyncio.sleep(random.random() * 3)

    start_server = websockets.serve(time, '192.168.4.177', 9876)

    asyncio.get_event_loop().run_until_complete(start_server)
    asyncio.get_event_loop().run_forever()

return render(request,"ws.html")

When the URL is handled by this view the above mentioned error occurs.

My ws.html is the exact copy from the above mentioned websockets documentation example

Could you please show us what you tried? – juankysmith Jul 07 '16 at 07:36 — juankysmith, Jul 07 '16 at 07:36
@juankysmith Please see the update. – arminrock Jul 07 '16 at 07:48 — arminrock, Jul 07 '16 at 07:48

score 1 · Answer 1 · answered Jul 07 '16 at 07:58

1

Django's request/response cycle is strictly synchronous. What you are trying to do is not possible in a normal Django view.

You might be interested in Django Channels, a project that aims to remove this limitation.

answered Jul 07 '16 at 07:58

jaap3

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score 1 · Answer 2 · answered Jul 07 '16 at 07:58

You can't really do this. I can't say why you're getting the exact errors you're getting, but a GET request to a Django view needs to return a response after some finite time, not run forever, otherwise the browser (or other parts in between like Nginx) will see the non-response as a timeout. If you want to run a websocket server, do it in a separate process outside of Django's.

There is much ongoing work to add async functionality and websockets to Django, in the form of channels -- I think the docs at http://channels.readthedocs.io/en/latest/ are the latest version of the code you can currently already use; hopefully it will be part of Django 1.10. Current version should be useable as a Django app that will allow you to make websockets in Django, but it's not as easy as your try above.

Note that Channels [will not be](https://groups.google.com/forum/#!topic/django-developers/QRd4v7OErT8) in 1.10, but will live as a separate app for the foreseeable future. That shouldn't stop you from using it, though! — knbk, Jul 07 '16 at 08:43
After the answers i started reading and learning the concept and implementation of ``channles``. Thanks for the feedbacks — arminrock, Jul 07 '16 at 09:15

Django websockets implementation

2 Answers2