python - How can I sort by the id of a ManyToManyField in Django?

Question

I've got a ManyToManyField in a user object and it's used to map the users that user is following. I'm trying to show a subset list of who they have most recently followed. Is there a trick in .order_by() that will allow me to sort by the id of the ManyToManyField? The data is there, right?

# (people the user is following)
following = models.ManyToManyField(User, related_name="following", blank=True)

theuser.following.filter(user__is_active=True).order_by("user__id")

That will give me a list of the users the user is following but ordered by when they joined. I want the order of the following list to be in order of when the user followed them.

Answer 1

I just found a way to do this without having to create a class for the relationship. It relies on the extra feature that lets you add additional columns to output. In your example it would look like:

theuser.following.filter(user__is_active=True)
    .extra(select={'creation_seq': 'appname_user_user_following.id'})
    .order_by("creation_seq")

Notice that appname_user_user_following is the name of the relationship table Django creates under the covers. It's deterministic and something you can get and set via meta-mechanisms, but it's pretty much safe to hardcode.

Here's an example of the SQL that's being created under the covers with fake table and columns names:

SELECT (appname_user_user_following.id) AS `creation_seq`, `appname_user`.`id`
FROM `appname_user` INNER JOIN `appname_user_user_following` ON
(`appname_user`.`id` = `appname_user_user_following`.`user_id`) WHERE
`appname_user_user_following`.`user_followed_id` = 1  ORDER BY `creation_seq` ASC';