Django - How to track if a user is online/offline in realtime?

问题

I'm considering to use django-notifications and Web Sockets to send real-time notifications to iOS/Android and Web apps. So I'll probably use Django Channels.

Can I use Django Channels to track online status of an user real-time? If yes then how I can achieve this without polling constantly the server?

I'm looking for a best practice since I wasn't able to find any proper solution.

UPDATE:

What I have tried so far is the following approach: Using Django Channels, I implemented a WebSocket consumer that on connect will set the user status to 'online', while when the socket get disconnected the user status will be set to 'offline'. Originally I wanted to included the 'away' status, but my approach cannot provide that kind of information. Also, my implementation won't work properly when the user uses the application from multiple device, because a connection can be closed on a device, but still open on another one; the status would be set to 'offline' even if the user has another open connection.

class MyConsumer(AsyncConsumer):

    async def websocket_connect(self, event):
        # Called when a new websocket connection is established
        print("connected", event)
        user = self.scope['user']
        self.update_user_status(user, 'online')

    async def websocket_receive(self, event):
        # Called when a message is received from the websocket
        # Method NOT used
        print("received", event)

    async def websocket_disconnect(self, event):
        # Called when a websocket is disconnected
        print("disconnected", event)
        user = self.scope['user']
        self.update_user_status(user, 'offline')

    @database_sync_to_async
    def update_user_status(self, user, status):
        """
        Updates the user `status.
        `status` can be one of the following status: 'online', 'offline' or 'away'
        """
        return UserProfile.objects.filter(pk=user.pk).update(status=status)

NOTE:

My current working solution is using the Django REST Framework with an API endpoint to let client apps send HTTP POST request with current status. For example, the web app tracks mouse events and constantly POST the online status every X seconds, when there are no more mouse events POST the away status, when the tab/window is about to be closed, the app sends a POST request with status offline. THIS IS a working solution, depending on the browser I have issues when sending the offline status, but it works.

What I'm looking for is a better solution that doesn't need to constantly polling the server.

回答1:

Using WebSockets is definitely the better approach.

Instead of having a binary "online"/"offline" status, you could count connections: When a new WebSocket connects, increase the "online" counter by one, when a WebSocket disconnects, decrease it. So that, when it is 0, then the user is offline on all devices.

Something like this

@database_sync_to_async
def update_user_incr(self, user):
    UserProfile.objects.filter(pk=user.pk).update(online=F('online') + 1)

@database_sync_to_async
def update_user_decr(self, user):
    UserProfile.objects.filter(pk=user.pk).update(online=F('online') - 1)

回答2:

The best approach is using Websockets.

But I think you should store not just the status, but also a session key or a device identification. If you use just a counter, you are losing valuable information, for example, from what device is the user connected at a specific moment. That is key in some projects. Besides, if something wrong happens (disconnection, server crashes, etc), you are not going to be able to track what counter is related with each device and probably you'll need to reset the counter at the end.

I recommend you to store this information in another related table:

from django.db import models
from django.conf import settings


class ConnectionHistory(models.Model):
    ONLINE = 'online'
    OFFLINE = 'offline'
    STATUS = (
        (ONLINE, 'On-line'),
        (OFFLINE, 'Off-line'),
    )
    user = models.ForeignKey(
        settings.AUTH_USER_MODEL,
        on_delete=models.CASCADE
    )
    device_id = models.CharField(max_lenght=100)
    status = models.CharField(
        max_lenght=10, choices=STATUS,
        default=ONLINE
    )
    first_login = models.DatetimeField(auto_now_add=True)
    last_echo = models.DatetimeField(auto_now=True)

    class Meta:
        unique_together = (("user", "device_id"),)

This way you have a record per device to track their status and maybe some other information like ip address, geoposition, etc. Then you can do something like (based on your code):

@database_sync_to_async
def update_user_status(self, user, device_id, status):
    return ConnectionHistory.objects.get_or_create(
        user=user, device_id=device_id,
    ).update(status=status)

How to get a device identification

There are plenty of libraries do it like https://www.npmjs.com/package/device-uuid. They simply use a bundle of browser parameters to generate a hash key. It is better than use session id alone, because it changes less frencuently.

Tracking away status

After each action, you can simply update last_echo. This way you can figured out who is connected or away and from what device.

Advantage: In case of crash, restart, etc, the status of the tracking could be re-establish at any time.

来源：https://stackoverflow.com/questions/51931038/django-how-to-track-if-a-user-is-online-offline-in-realtime

标签

django

django-channels

django-notification