I try something with readonly field in Django 1.8.7, let say I have some code like the following:
class MyAdmin(admin.ModelAdmin):
readonly_fields = ('a', 'b')
def get_readonly_fields(self, request, obj=None):
if not request.user.is_superuser:
self.readonly_fields += ('c')
return super(MyAdmin, self).get_readonly_fields(request, obj)
first I login with super admin and access that admin page change_form,
the code is works well, then I login with staff user, then still works well, again, I try login with superadmin, but the read only fields rendered is for the non-superadmin user,
again I clear the browser cache, try again with super admin, but still not work correctly. I try restart the server, then it work normally until I repeat the same step above I do, this weird thing come again.
Anyone know why this happen ? I think this is looks like some bug but not sure.
Thanks in Advance.
The bug is not in Django, but in your code. In your get_readonly_fields method you modify the readonly_fields attribute; those modifications persist, since the admin object lives for the lifetime of the process.
Don't do that. get_readonly_fields is supposed to return a value, not modify the attribute. Just do:
def get_readonly_fields(self, request, obj=None):
rfo = super(MyAdmin, self).get_readonly_fields(request, obj)
if not request.user.is_superuser:
rfo += ('c')
return rfo
Related
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):
i am really new to Django, but since i am doing a fairly easy app i was suggested by a friend to use only the admin site which is fast and easy, i am almost done with my project, but i need one thing i don't seem to find anywhere.
My model is about adding programming problems, however i need to have a field that identifies which user added it and that only that user and the super users can erase or change the problem he just added.
Of course adding the field is quite simple, however how do i recognize who is adding the problem? and how to validate only him and the super users can change or delete that said problem?. I believe this is the most challenging phase of this project, can you help me?.
Thanks in advance :)!
EDIT: this is what i tried after the answer i just got recently, but i am quite stuck :/
EDIT2: this is now how it looks, but it gives my a type error and brings down the whole adminsite with this :
"has_change_permission() takes exactly 3 arguments (2 given)"
EDIT3: now i have changed the code but it still let regular staff users to erase delete entries, and sadly now it gives the error in EDIT 2 when i try to modify the content of the entry.
EDIT4: Finally thanks to your help i could make it impossible for a common user to delete directly on the database, however the default delete by queryset still works for them, what can i do?
EDIT5: Thanks for all the help you gave me today, now it works and it is wonderful! i leave the code so someone else with this same issue can just do it, i had to do such a long research to get rid of the "delete_element" from django's admin the one that is setted as default.
from django.contrib import admin
from .models import Problemas
# Register your models here.
from django.contrib.admin.actions import delete_selected as delete_selected_
def delete_selected(modeladmin,request,queryset):
for obj in queryset:
if (obj.User==request.user.username or request.user.is_superuser):
obj.delete()
class ProblemasAdmin(admin.ModelAdmin):
list_display = ('Juez', 'Nombre', 'Categoria','Dificultad','URL','User')
list_filter = ('Juez','Categoria','Dificultad','User')
search_fields = ['Nombre']
readonly_fields = ('User',)
list_per_page = 20
actions = [delete_selected]
def save_model(self, request, obj, form, change):
if (obj.User==""):
obj.User = request.user.username
obj.save()
def delete_model(self,request,obj):
for o in obj.all():
if (o.User==request.user.username or request.user.is_superuser):
o.delete()
def has_change_permission(self,request,obj=None):
return obj==None or request.user.username == obj.User or request.user.is_superuser
def has_delete_permission(self,request,obj=None):
return obj==None or request.user.username == obj.User or request.user.is_superuser
admin.site.register(Problemas,ProblemasAdmin)
#register(QuestionModel)
class QuestionModelAdmin(ModelAdmin):
def has_change_permission(self,request,obj=None):
return request.user == obj.owner
def has_delete_permission(self,request,obj=None):
return request.user == obj.owner
I think should work
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.
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)
One of the functionalities in a Django project I am writing is sending a newsletter. I have a model, Newsletter and a function, send_newsletter, which I have registered to listen to Newsletter's post_save signal. When the newsletter object is saved via the admin interface, send_newsletter checks if created is True, and if yes it actually sends the mail.
However, it doesn't make much sense to edit a newsletter that has already been sent, for the obvious reasons. Is there a way of making the Newsletter object read-only once it has been saved?
Edit:
I know I can override the save method of the object to raise an error or do nothin if the object existed. However, I don't see the point of doing that. As for the former, I don't know where to catch that error and how to communicate the user the fact that the object wasn't saved. As for the latter, giving the user false feedback (the admin interface saying that the save succeded) doesn't seem like a Good Thing.
What I really want is allow the user to use the Admin interface to write the newsletter and send it, and then browse the newsletters that have already been sent. I would like the admin interface to show the data for sent newsletters in an non-editable input box, without the "Save" button. Alternatively I would like the "Save" button to be inactive.
You can check if it is creation or update in the model's save method:
def save(self, *args, **kwargs):
if self.pk:
raise StandardError('Can\'t modify bla bla bla.')
super(Payment, self).save(*args, **kwargs)
Code above will raise an exception if you try to save an existing object. Objects not previously persisted don't have their primary keys set.
Suggested reading: The Zen of Admin in chapter 17 of the Django Book.
Summary: The admin is not designed for what you're trying to do :(
However, the 1.0 version of the book covers only Django 0.96, and good things have happened since.
In Django 1.0, the admin site is more customizable. Since I haven't customized admin myself, I'll have to guess based on the docs, but I'd say overriding the model form is your best bet.
use readonlyadmin in ur amdin.py.List all the fields which u want to make readonly.After creating the object u canot edit them then
use the link
http://www.djangosnippets.org/snippets/937/
copy the file and then import in ur admin.py and used it
What you can easily do, is making all fields readonly:
class MyModelAdmin(ModelAdmin):
form = ...
def get_readonly_fields(self, request, obj=None):
if obj:
return MyModelAdmin.form.Meta.fields
else: # This is an addition
return []
As for making the Save disappear, it would be much easier if
has_change_permission returning False wouldnt disable even displaying the form
the code snippet responsible for rendering the admin form controls (the admin_modify.submit_row templatetag), wouldnt use show_save=True unconditionally.
Anyways, one way of making that guy not to be rendered :
Create an alternate version of has_change_permission, with proper logic:
class NoSaveModelAdminMixin(object):
def render_change_form(self, request, context, add=False, change=False, form_url='', obj=None):
response = super(NoSaveModelAdmin, self).render_change_form(request, context, add, change,form_url, obj)
response.context_data["has_change_permission"] = self.has_real_change_permission(request, obj)
def has_real_change_permission(self, request, obj):
return obj==None
def change_view(self, request, object_id, extra_context=None):
obj = self.get_object(request, unquote(object_id))
if not self.has_real_change_permission(request, obj) and request.method == 'POST':
raise PermissionDenied
return super(NoSaveModelAdmin, self).change_view(request, object_id, extra_context=extra_context)
Override the submit_row templatetag similar to this:
#admin_modify.register.inclusion_tag('admin/submit_line.html', takes_context=True)
def submit_row(context):
...
'show_save': context['has_change_permission']
...
admin_modify.submit_row = submit_row