Django user audit

Question

I would like to create a view with a table that lists all changes (created/modified) that a user has made on/for any object.

The Django Admin site has similar functionality but this only works for objects created/altered in the admin.

All my models have, in addition to their specific fields, following general fields, that should be used for this purpose:

created_by = models.ForeignKey(User, verbose_name='Created by', related_name='%(class)s_created_items',)
modified_by = models.ForeignKey(User, verbose_name='Updated by', related_name='%(class)s_modified_items', null=True)
created = CreationDateTimeField(_('created'))
modified = ModificationDateTimeField(_('modified'))

I tried playing around with:

u = User.objects.get(pk=1)
u.myobject1_created_items.all()
u.myobject1_modified_items.all()
u.myobject2_created_items.all()
u.myobject2_modified_items.all()
... # repeat for >20 models

...and then grouping them together with itertool's chain(). But the result is not a QuerySet which makes it kind of non-Django and more difficult to handle.

I realize there are packages available that will do this for me, but is it possible to achieve what I want using the above models, without using external packages? The required fields (created_by/modified_by and their timefields) are in my database already anyway.

Any idea on the best way to handle this?

score 1 · Answer 1 · answered Feb 23 '15 at 15:46

Django admin uses generic foreign keys to handle your case so you should probably do something like that. Let's take a look at how django admn does it (https://github.com/django/django/blob/master/django/contrib/admin/models.py):

class LogEntry(models.Model):
    action_time = models.DateTimeField(_('action time'), auto_now=True)
    user = models.ForeignKey(settings.AUTH_USER_MODEL)
    content_type = models.ForeignKey(ContentType, blank=True, null=True)
    object_id = models.TextField(_('object id'), blank=True, null=True)
    object_repr = models.CharField(_('object repr'), max_length=200)
    action_flag = models.PositiveSmallIntegerField(_('action flag'))
    change_message = models.TextField(_('change message'), blank=True)

So, you can add an additional model (LogEntry) that will hold a ForeignKey to the user that changed (added / modified) the object and a GenericForeignKey (https://docs.djangoproject.com/en/1.7/ref/contrib/contenttypes/#generic-relations) to the object that was modified.

Then, you can modify your views to add LogEntry objects when objects are modified. When you want to display all changes by a User, just do something like:

user = User.objects.get(pk=1)
changes = LogEntry.objects.filter(user=user)
# Now you can use changes for your requirement!

I've written a nice blog post about that (auditing objects in django) which could be useful: http://spapas.github.io/2015/01/21/django-model-auditing/#adding-simple-auditing-functionality-ourselves

Django user audit

1 Answers1