Django: DRY principle and UserPassesTestMixin - python

I have a model named Post and have a field there called owner (foreign key to User). Of course, only owners can update or delete their own posts.
That being said, I use login_required decorator in the views to make sure the user is logged in but then, I also need to make sure the user trying to update/delete the question is the owner.
As I'm using Django: Generic Editing Views the documentation says I need to use Django: UserPassesTestMixin.
This validation will be done for the update and delete views. DRY, what is the way to go about this? should I create a class named TestUserOwnerOfPost and create a test_func() and then make the update and delete views inherit from it?
Cause that's what I have tried and didn't work, code below:
from django.views.generic.edit import UpdateView
from django.contrib.auth.decorators import login_required
from django.contrib.auth.mixins import UserPassesTestMixin
class TestUserOwnerOfPost(UserPassesTestMixin):
def test_func(self):
return self.request.user == self.post.owner
class EditPost(UpdateView, TestUserOwnerOfPost):
model = Post
#method_decorator(login_required)
def dispatch(self, *args, **kwargs):
return super(EditPost, self).dispatch(*args, **kwargs)
With the code above, every logged-in user in the system can edit/delete any post. What am I doing wrong? am I missing something? thanks.

The first problem is that the order of the classes you inherit is incorrect, as #rafalmp says.
However, fixing that doesn't solve the problem, because the UserPassesTest mixin performs the test before running the view. This means that it's not really suitable to check the owner of self.object, because self.object has not been set yet. Note I'm using self.object instead of self.post -- I'm don't think that the view ever sets self.post but I might be wrong about that.
One option is to call self.get_object() inside the test function. This is a bit inefficient because your view will fetch the object twice, but in practice it probably doesn't matter.
def test_func(self):
self.object = self.get_object()
return self.request.user == self.object.owner
Another approach is to override get_queryset, to restrict it to objects owned by the user. This means the user will get a 404 error if they do not own the object. This behaviour is not exactly the same as the UserPassesTestMixin, which will redirect to a login page, but it might be ok for you.
class OwnerQuerysetMixin(object):
def get_queryset(self):
queryset = super(OwnerQuerysetMixin, self).get_queryset()
# perhaps handle the case where user is not authenticated
queryset = queryset.filter(owner=self.request.user)
return queryset

The order of the classes you inherit from matters. For your access control to work, it must be enforced before UpdateView is executed:
class EditPost(TestUserOwnerOfPost, UpdateView):

Related

Django get custom queryset into ListView

I have a listview that I access in a pretty bog standard way to return all metaobjects.
#url
url(r'^metaobject/$', MetaObjectList.as_view(),name='metaobject_list'),
#ListView
class MetaObjectList(ListView):
model = MetaObject
I've recently added a search form that I want to scan my objects (I've got about 5 fields but I've simplified the example). What I'd like to do is re-use my MetaObjectList class view with my specific subset. I am guessing I need to override the get_queryset method but I'm not clear in how I get the queryset from my FormView into the listview. I mucked around a bit with calling the as_view() in the formveiw's form_valid function with additional parameters but couldn't get it to work and it seemed hacky anyway.
class SearchView(FormView):
template_name = 'heavy/search.html'
form_class = SearchForm
#success_url = '/thanks/'
def form_valid(self, form):
#build a queryset based on form
searchval=form.cleaned_data['search']
list = MetaObject.objects.filter(val=search)
#where to from here?
I also looked at trying to post the data from the form view over to the listview but that seemed like I'd need to re-write the form logic into the listview.
I'm on python 3.x and django 1.11.
I found what I feel is more elegant than the comment on the question:
My form valid now points to the list object's as_view method and passes the request and the queryset I want
def form_valid(self, form):
#build a queryset based on form
searchval=form.cleaned_data['search']
list = MetaObject.objects.filter(val=search)
return MetaObjectList.as_view()(self.request,list)
This hits the ListView as a post which I use to alter the queryset
class MetaObjectList(ListView):
model = MetaObject
queryset = MetaObject.objects.prefetch_related('object_type','domain')
def post(self, request, *args, **kwargs):
self.queryset = args[0]
return self.get(request, *args, **kwargs)
The only obvious change is using kwargs to make it a bit clearer. Otherwise this seems to work well.

Django mixin not working as expected

I want to prevent logged-in users to access login and register forms.
I've build custom mixin, but it isn't working. The problem is that even if the user is logged in, he can access login and register forms instead of beeing redirected to homepage.
My Mixin
class MustBeAnonymousMixin(object):
''' Only anonymous users (not logged in) may access login and register
'''
def dispath(self, *args, **kwargs):
if not self.request.user.is_anonymous:
return redirect(reverse('homepage'))
return super(MustBeAnonymousMixin, self).dispatch(*args, **kwargs)
LoginFormView
class LoginFormView(MustBeAnonymousMixin, TemplateView):
'''
Display basic user login form
'''
template_name = 'members/login.html'
def get_context_data(self, **kwargs):
context = super(LoginFormView, self).get_context_data(**kwargs)
context['login_form'] = UserLoginForm()
return context
I'm using Django 1.8. What am I doing wrong?
For another case where mixin does not work:
Remember: "Mixin param" must stand before "GenericView param"
Correct:
class PostDelete(LoginRequiredMixin, generic.DeleteView):
Incorrect:
class PostDelete(generic.DeleteView, LoginRequiredMixin):
Fix the typo in dispath and use is_authenticated() instead of is_anonymous (as indicated in the previous answer already)
is_anonymous should be a function call, and you probably should not use it:
is_anonymous()
Always returns False. This is a way of differentiating User and
AnonymousUser objects. Generally, you should prefer using is_authenticated() to this method.

How to prevent user changing URL <pk> to see other submission data Django

I'm new to the web development world, to Django, and to applications that require securing the URL from users that change the foo/bar/pk to access other user data.
Is there a way to prevent this? Or is there a built-in way to prevent this from happening in Django?
E.g.:
foo/bar/22 can be changed to foo/bar/14 and exposes past users data.
I have read the answers to several questions about this topic and I have had little luck in an answer that can clearly and coherently explain this and the approach to prevent this. I don't know a ton about this so I don't know how to word this question to investigate it properly. Please explain this to me like I'm 5.
There are a few ways you can achieve this:
If you have the concept of login, just restrict the URL to:
/foo/bar/
and in the code, user=request.user and display data only for the logged in user.
Another way would be:
/foo/bar/{{request.user.id}}/
and in the view:
def myview(request, id):
if id != request.user.id:
HttpResponseForbidden('You cannot view what is not yours') #Or however you want to handle this
You could even write a middleware that would redirect the user to their page /foo/bar/userid - or to the login page if not logged in.
I'd recommend using django-guardian if you'd like to control per-object access. Here's how it would look after configuring the settings and installing it (this is from django-guardian's docs):
>>> from django.contrib.auth.models import User
>>> boss = User.objects.create(username='Big Boss')
>>> joe = User.objects.create(username='joe')
>>> task = Task.objects.create(summary='Some job', content='', reported_by=boss)
>>> joe.has_perm('view_task', task)
False
If you'd prefer not to use an external library, there's also ways to do it in Django's views.
Here's how that might look:
from django.http import HttpResponseForbidden
from .models import Bar
def view_bar(request, pk):
bar = Bar.objects.get(pk=pk)
if not bar.user == request.user:
return HttpResponseForbidden("You can't view this Bar.")
# The rest of the view goes here...
Just check that the object retrieved by the primary key belongs to the requesting user. In the view this would be
if some_object.user == request.user:
...
This requires that the model representing the object has a reference to the User model.
In my project, for several models/tables, a user should only be able to see data that he/she entered, and not data that other users entered. For these models/tables, there is a user column.
In the list view, that is easy enough to implement, just filter the query set passed to the list view for model.user = loggged_id.user.
But for the detail/update/delete views, seeing the PK up there in the URL, it is conceivable that user could edit the PK in the URL and access another user's row/data.
I'm using Django's built in class based views.
The views with PK in the URL already have the LoginRequiredMixin, but that does not stop a user from changing the PK in the URL.
My solution: "Does Logged In User Own This Row Mixin"
(DoesLoggedInUserOwnThisRowMixin) -- override the get_object method and test there.
from django.core.exceptions import PermissionDenied
class DoesLoggedInUserOwnThisRowMixin(object):
def get_object(self):
'''only allow owner (or superuser) to access the table row'''
obj = super(DoesLoggedInUserOwnThisRowMixin, self).get_object()
if self.request.user.is_superuser:
pass
elif obj.iUser != self.request.user:
raise PermissionDenied(
"Permission Denied -- that's not your record!")
return obj
Voila!
Just put the mixin on the view class definition line after LoginRequiredMixin, and with a 403.html template that outputs the message, you are good to go.
In django, the currently logged in user is available in your views as the property user of the request object.
The idea is to filter your models by the logged in user first, and then if there are any results only show those results.
If the user is trying to access an object that doesn't belong to them, don't show the object.
One way to take care of all of that is to use the get_object_or_404 shortcut function, which will raise a 404 error if an object that matches the given parameters is not found.
Using this, we can just pass the primary key and the current logged in user to this method, if it returns an object, that means the primary key belongs to this user, otherwise it will return a 404 as if the page doesn't exist.
Its quite simple to plug it into your view:
from django.shortcuts import get_object_or_404, render
from .models import YourModel
def some_view(request, pk=None):
obj = get_object_or_404(YourModel, pk=pk, user=request.user)
return render(request, 'details.html', {'object': obj})
Now, if the user tries to access a link with a pk that doesn't belong to them, a 404 is raised.
You're going to want to look into user authentication and authorization, which are both supplied by [Django's Auth package] (https://docs.djangoproject.com/en/4.0/topics/auth/) . There's a big difference between the two things, as well.
Authentication is making sure someone is who they say they are. Think, logging in. You get someone to entire their user name and password to prove they are the owner of the account.
Authorization is making sure that someone is able to access what they are trying to access. So, a normal user for instance, won't be able to just switch PK's.
Authorization is well documented in the link I provided above. I'd start there and run through some of the sample code. Hopefully that answers your question. If not, hopefully it provides you with enough information to come back and ask a more specific question.
This is a recurring question and also implies a serious security flaw. My contribution is this:
There are 2 basic aspects to take care of.
The first is the view:
a) Take care to add a decorator to the function-based view (such as #login_required) or a mixin to the class-based function (such as LoginRequiredMixin). I find the official Django documentation quite helpful on this (https://docs.djangoproject.com/en/4.0/topics/auth/default/).
b) When, in your view, you define the data to be retrieved or inserted (GET or POST methods), the data of the user must be filtered by the ID of that user. Something like this:
def get(self, request, *args, **kwargs):
self.object = self.get_object(queryset=User.objects.filter(pk=self.request.user.id))
return super().get(request, *args, **kwargs)
The second aspect is the URL:
In the URL you should also limit the URL to the pk that was defined in the view. Something like this:
path('int:pk/blog-add/', AddBlogView.as_view(), name='blog-add'),
In my experience, this prevents that an user sees the data of another user, simply by changing a number in the URL.
Hope it helps.
In django CBV (class based views) you can prevent this by comparing the
user entered pk and the current logged in user:
Note: I tested it in django 4 and python 3.9.
from django.http import HttpResponseForbidden
class UserDetailView(LoginRequiredMixin, DetailView):
model = your_model
def dispatch(self, request, *args, **kwargs):
if kwargs.get('pk') != self.request.user.pk:
return HttpResponseForbidden(_('You do not have permission to view this page'))
return super().dispatch(request, *args, **kwargs)

Django multi model best practice

Basically what I'm trying to achieve is a multi-model django app where different models take advantage of the same views. For example I've got the models 'Car' 'Make' 'Model' etc and I want to build a single view to perform the same task for each, such as add, delete and edit, so I don't have to create a seperate view for add car, ass make etc. I've built a ModelForm and Model object for each and would want to create a blank object when adding and a pre-populated form object when editing (through the form instance arg), with objects being determined via url parameters.
Where I'm stuck is that I'm not sure what the best way to so this is. At the moment I'm using a load of if statements to return the desired object or form based on parameters I'm giving it, which get's a bit tricky when different forms need specifying and whether they need an instance or not. Although this seems to be far from the most efficient way of achieving this.
Django seems to have functions to cover most repetitive tasks, is there some magic I'm missing here?
edit - Here's an example of what I'm doing with the arguments I'm passing into the url:
def edit_object(request, object, id):
if(object==car):
form = carForm(instance = Car.objects.get(pk=id)
return render(request, 'template.html', {'form':form})
What about using Class Based Views? Using CBVs is the best way in Django to make reusable code. For this example maybe it can be a little longer than function based views, but when the project grows up it makes the difference. Also remember "Explicit is better than implicit".
urls.py
# Edit
url(r'^car/edit/(?P<pk>\d+)/$', EditCar.as_view(), name='edit-car'),
url(r'^make/edit/(?P<pk>\d+)/$', EditMake.as_view(), name='edit-make'),
# Delete
url(r'^car/delete/(?P<pk>\d+)/$', DeleteCar.as_view(), name='delete-car'),
url(r'^make/delete/(?P<pk>\d+)/$', DeleteMake.as_view(), name='delete-make'),
views.py
class EditSomethingMixin(object):
"""Use Mixins to reuse common behavior"""
template_name = 'template-edit.html'
class EditCar(EditSomethingMixin, UpdateView):
model = Car
form_class = CarForm
class EditMake(EditSomethingMixin, UpdateView):
model = Make
form_class = MakeForm
class DeleteSomethingMixin(object):
"""Use Mixins to reuse common behavior"""
template_name = 'template-delete.html'
class DeleteCar(DeleteSomethingMixin, DeleteView):
model = Car
class DeleteMake(DeleteSomethingMixin, DeleteView):
model = Make
Just pass your class and form as args to the method then call them in the code.
def edit_object(request, model_cls, model_form, id):
form = model_form(instance = model_cls.objects.get(pk=id)
return render(request, 'template.html', {'form':form})
then just pass in the correct classes and forms in your view methods
def edit_car(request,id):
return edit_object(request, Car, CarForm, id)
each method knows what classes to pass, so you eliminate the if statements.
urls.py
url(r'^car/delete/(?<pk>\d+)/', edit, {'model': Car})
url(r'^make/delete/(?<pk>\d+)/', edit, {'model': Make})
views.py
def edit(request, id, model):
model.objects.get(id=id).delete()

Django, class-views: How can I save session data with a form's object?

I'm trying to store the username from the current request's session into a db object. How can I do this from within a class-based view? Is there a "clean" way to do this? What should I override/subclass?
I have a model that looks like this:
from django.contrib.auth.models import User
class Entry(django.db.models.Model):
...
author = models.ForeignKey(User, editable=False)
I also have a view based on the built-in generic view django.views.generic.CreateView. I'm also using the default ModelForm class that goes with my model, and the default {{ form }} in my template. AFAIK, the session and authentication apps/middleware are set up properly---as per default in new Django projects.
I found this post, which is getting at about the same thing, but from the wrong angle, and using function views instead.
My thinking so far was to override something in the form class and insert the username into the cleaned data. Is there a better way? Is there a right way?
Edit: Solution so far, non-working, with an IntegrityError: author_id cannot be null
from django.views.generic import CreateView
class Index(CreateView):
model = magicModel
template_name = "index.html"
success_url = "/magicWorked"
...
def form_valid(self, form):
self.object = form.save(commit=False)
self.object.author = request.user
return super(Index, self).form_valid(form)
I wrote this based on what I found in django/views/generic/edit.py, which uses this implementation for class ModelFormMixin:
def form_valid(self, form):
self.object = form.save()
return super(ModelFormMixin, self).form_valid(form)
This is the method called by super().form_valid() above.
Edit: The problem with my solution was my understanding of Python's inheritance model. When the super-class calls form_valid(), it calls its own version, not my override; my code was never running at all.
The "correct" way to do this is to write your own view for object creation if the generic view doesn't suffice. Creation views are relatively short and there are numerous examples of how to save foreign keys.
Incidentally, Django's 1.3 docs say somewhere in there that modifications to the authentication model used by the admin app are being "discussed," such as adding per-instance permissions. (The current auth model supports only per model permissions.) The dev's might also add an implementation for what I'm trying to achieve. After all, user-associated data is used by nearly all websites.

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