I have the following method:
def _attempt(actor):
if actor.__class__ != User:
raise TypeError
Which is called from a view:
self.object.attempt(self.request.user)
As you can see, the _attempt method expects actor to be type django.contrib.auth.models.User
, however the object appears to be of type django.utils.functional.SimpleLazyObject
. Why is this so? And more importantly, how can I convert the LazyObject
(which apparently is a kind of wrapper for a User object) into a User
object?
More info on Request.user
is available here: https://docs.djangoproject.com/en/dev/ref/request-response/#django.http.HttpRequest.user This documentation seems to indicate the request.user
should be a User
object...
======Post-edit=====
I've got the following method now:
def _attempt(obj, action, actor, msg):
actor.is_authenticated()
if isinstance(actor, LazyObject):
print type(actor)
I'm passing a user, however the if
condition is still true, actor is still a LazyObject
. Why is this so?
See my answer on a similar question.
Django lazy loads request.user
so that it can be either User
or AnonymousUser
depending on the authentication state. It only "wakes up" and returns the appropriate class when an attribute is accessed on it. Unfortunately, __class__
doesn't count because that's a primitive class attribute. There's occasions where you might need to know that this is actually a SimpleLazyObject
type, and therefore it would be wrong to proxy it on to User
or AnonymousUser
.
Long and short, you simply can't do this comparison as you have it. But, what are you really trying to achieve here? If you're trying to check if it's a User
or AnonymousUser
, there's request.user.is_authenticated()
for that, for example.
As a general rule though, you shouldn't abuse duck typing. A parameter should always be a particularly type or subtype (User
or UserSubClass
), even though it doesn't have to be. Otherwise, you end up with confusing and brittle code.