如何从Django Channels web socket包获取当前用户？

Question

我在学习这个教程：Finally, Real-Time Django Is Here: Get Started with Django Channels。

我想使用Django用户对象而不是handle变量来扩展应用程序。但是如何在我的ws_recieve(message)函数中从收到的WebSocket包中获取当前用户？

我注意到web socket包中的csrftoken和sessionid的前十位数字都匹配一个正常的HTTP请求。我可以用这些信息获取当前用户吗？

作为参考，收到的数据包如下所示：

{'channel': <channels.channel.Channel object at 0x110ea3a20>,
 'channel_layer': <channels.asgi.ChannelLayerWrapper object at 0x110c399e8>,
 'channel_session': <django.contrib.sessions.backends.db.SessionStore object at 0x110d52cc0>,
 'content': {'client': ['127.0.0.1', 52472],
             'headers': [[b'connection', b'Upgrade'],
                         [b'origin', b'http://0.0.0.0:8000'],
                         [b'cookie',
                          b'csrftoken=EQLI0lx4SGCpyTWTJrT9UTe1mZV5cbNPpevmVu'
                          b'STjySlk9ZJvxzHj9XFsJPgWCWq; sessionid=kgi57butc3'
                          b'zckszpuqphn0egqh22wqaj'],
                         [b'cache-control', b'no-cache'],
                         [b'sec-websocket-version', b'13'],
                         [b'sec-websocket-extensions',
                          b'x-webkit-deflate-frame'],
                         [b'host', b'0.0.0.0:8000'],
                         [b'upgrade', b'websocket'],
                         [b'sec-websocket-key', b'y2Lmb+Ej+lMYN+BVrSXpXQ=='],
                         [b'user-agent',
                          b'Mozilla/5.0 (Macintosh; Intel Mac OS X 10_11_6) '
                          b'AppleWebKit/602.1.50 (KHTML, like Gecko) Version'
                          b'/10.0 Safari/602.1.50'],
                         [b'pragma', b'no-cache']],
             'order': 0,
             'path': '/chat-message/',
             'query_string': '',
             'reply_channel': 'websocket.send!UZaOWhupBefN',
             'server': ['127.0.0.1', 8000]},
 'reply_channel': <channels.channel.Channel object at 0x110ea3a90>}

Answer 1

注意：这个问题是特定于Channels 1.x和Django会话系统的，如果您是以Channels 2.x的方式来这里的，请从下面的链接中阅读文档，他们非常清楚如何做到这一点，Channels 1.x文档让一些人（包括我）感到困惑，这就是为什么要提出这个问题和其他问题，Channels 2.x文档是清楚地说明如何实现这一点：
https://channels.readthedocs.io/en/latest/topics/authentication.html#django-authentication

---

频道1.x:

通过更改consumers.py中的装饰符以匹配文档，您可以访问user和http_session属性：

You get access to a user’s normal Django session using the http_session decorator - that gives you a message.http_session attribute that behaves just like request.session. You can go one further and use http_session_user which will provide a message.user attribute as well as the session attribute.

因此示例中的^{}文件应该变成以下内容：

from channels.auth import http_session_user, channel_session_user, channel_session_user_from_http  

@channel_session_user_from_http  
def ws_connect(message):  
    ...  
    ...


@channel_session_user  
def ws_receive(message):  
    # You can check for the user attr like this  
    log.debug('%s', message.user)  
    ...  
    ...  


@channel_session_user  
def ws_disconnect(message):  
    ...  
    ...  

注意装饰器的变化
另外，我建议你不要在这个例子上做任何事情

有关详细信息，请参见：https://channels.readthedocs.io/en/1.x/getting-started.html#authentication

Answer 2

2018年更新答案via the docs：

To access the user, just use self.scope["user"] in your consumer code:

class ChatConsumer(WebsocketConsumer):

    def connect(self, event):
        self.user = self.scope["user"]

    def receive(self, event):
        username_str = None
        username = self.scope["user"]
        if(username.is_authenticated()):
            username_str = username.username
            print(type(username_str))
            #pdb.set_trace() # optional debugging