[Answer]-Django user audit

by

1👍

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

👤Serafeim

Leave a comment