Django 3.0 + каналы + ASGI + TokenAuthMiddleware

Question

Я обновился до Django 3.0 и теперь получаю эту ошибку при использовании websockets + TokenAuthMiddleware:

SynchronousOnlyOperation
You cannot call this from an async context - use a thread or sync_to_async.

Answer 1

Проблема в том, что вы не можете получить доступ к синхронному коду из асинхронного контекста. ВотTokenAuthMiddleware для Django 3.0:

# myproject.myapi.utils.py
from channels.auth import AuthMiddlewareStack
from channels.db import database_sync_to_async
from django.contrib.auth.models import AnonymousUser

from rest_framework.authtoken.models import Token


@database_sync_to_async
def get_user(headers):
    try:
        token_name, token_key = headers[b'authorization'].decode().split()
        if token_name == 'Token':
            token = Token.objects.get(key=token_key)
            return token.user
    except Token.DoesNotExist:
        return AnonymousUser()


class TokenAuthMiddleware:

    def __init__(self, inner):
        self.inner = inner

    def __call__(self, scope):
        return TokenAuthMiddlewareInstance(scope, self)


class TokenAuthMiddlewareInstance:
    """
    Yeah, this is black magic:
    https://github.com/django/channels/issues/1399
    """
    def __init__(self, scope, middleware):
        self.middleware = middleware
        self.scope = dict(scope)
        self.inner = self.middleware.inner

    async def __call__(self, receive, send):
        headers = dict(self.scope['headers'])
        if b'authorization' in headers:
            self.scope['user'] = await get_user(headers)
        inner = self.inner(self.scope)
        return await inner(receive, send)


TokenAuthMiddlewareStack = lambda inner: TokenAuthMiddleware(AuthMiddlewareStack(inner))

Используйте это так:

# myproject/routing.py
from myapi.utils import TokenAuthMiddlewareStack
from myapi.websockets import WSAPIConsumer

application = ProtocolTypeRouter({
    "websocket": TokenAuthMiddlewareStack(
        URLRouter([
            path("api/v1/ws", WSAPIConsumer),
        ]),
    ),

})
application = SentryAsgiMiddleware(application)

Answer 2

Как заявил @tapion, это решение больше не работает, так как каналы 3.x

Новое решение можно немного подправить:

class TokenAuthMiddleware:
    def __init__(self, inner):
        self.inner = inner

    async def __call__(self, scope, receive, send):
        headers = dict(scope['headers'])
        if b'authorization' in headers:
            scope['user'] = await get_user(headers)
        return await self.inner(scope, receive, send)