I'm using Django for an application, and wondering about an option in the admin.
Is it possible for django admin to redirect to the details page of an object, if only one exists in the list view?
For example,
if only this object exists:
redirect immediately to the change view on this object, without needing the user to click on the object.
I'm not using any custom view. I couldn't find any solution after 2 hours of search.
Thanks!
You can try this
def changelist_view(self, request, extra_context=None):
if self.model.objects.all().count() == 1:
obj = self.model.objects.all()[0]
return HttpResponseRedirect(reverse("admin:%s_%s_change" %(self.model._meta.app_label, self.model._meta.model_name), args=(obj.id,)))
return super(ItemAdmin, self).changelist_view(request=request, extra_context=extra_context)
Also check changelist_view parameters based on django version.
I found that in order to display in Django admin viewer application product items that have vendor set to null, one should visit:
/admin/viewer/product/?vendor__isnull=True
However, I am not sure how to enable that from the user interface. I would like to allow users to filter by null value of this field, but creating a custom Django action that returns HttpResponseRedirect is not a good option because Django actions require some elements to be selected.
Do I have any other elegant options that would require little or none modifications to Django admin template that would allow me to redirect the user to this specific URL? The two places where I thought this would fit is the navigation bar at the top of the page next to the "search" button or the drop-down menu at the top-right corner where user can log out and change password. Is there an interface where I could inject new options/views there?
You shouldn't do this via the URL. You should modify the queryset for the admin so that it returns the objects you want.
class MyModelAdmin(admin.ModelAdmin):
def get_queryset(self):
qs = super(MyModelAdmin, self).get_queryset()
if some_condition: # not sure what it depends on - maybe request.user?
qs = qs.filter(vendor__isnull=True)
return qs
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)
I have 2 models - for example, Book and Page.
Page has a foreign key to Book.
Each page can be marked as "was_read" (boolean), and I want to prevent deleting pages that were read (in the admin).
In the admin - Page is an inline within Book (I don't want Page to be a standalone model in the admin).
My problem - how can I achieve the behavior that a page that was read won't be deleted?
I'm using Django 1.4 and I tried several options:
Override "delete" to throw a ValidationError - the problem is that the admin doesn't "catch" the ValidationError on delete and you get an error page, so this is not a good option.
Override in the PageAdminInline the method - has_delete_permission - the problem here -it's per type so either I allow to delete all pages or I don't.
Are there any other good options without overriding the html code?
Thanks,
Li
The solution is as follows (no HTML code is required):
In admin file, define the following:
from django.forms.models import BaseInlineFormSet
class PageFormSet(BaseInlineFormSet):
def clean(self):
super(PageFormSet, self).clean()
for form in self.forms:
if not hasattr(form, 'cleaned_data'):
continue
data = form.cleaned_data
curr_instance = form.instance
was_read = curr_instance.was_read
if (data.get('DELETE') and was_read):
raise ValidationError('Error')
class PageInline(admin.TabularInline):
model = Page
formset = PageFormSet
You could disable the delete checkbox UI-wise by creating your own custom
formset for the inline model, and set can_delete to False there. For
example:
from django.forms import models
from django.contrib import admin
class MyInline(models.BaseInlineFormSet):
def __init__(self, *args, **kwargs):
super(MyInline, self).__init__(*args, **kwargs)
self.can_delete = False
class InlineOptions(admin.StackedInline):
model = InlineModel
formset = MyInline
class MainOptions(admin.ModelAdmin):
model = MainModel
inlines = [InlineOptions]
Another technique is to disable the DELETE checkbox.
This solution has the benefit of giving visual feedback to the user because she will see a grayed-out checkbox.
from django.forms.models import BaseInlineFormSet
class MyInlineFormSet(BaseInlineFormSet):
def add_fields(self, form, index):
super().add_fields(form, index)
if some_criteria_to_prevent_deletion:
form.fields['DELETE'].disabled = True
This code leverages the Field.disabled property added in Django 1.9. As the documentation says, "even if a user tampers with the field’s value submitted to the server, it will be ignored in favor of the value from the form’s initial data," so you don't need to add more code to prevent deletion.
In your inline, you can add the flag can_delete=False
EG:
class MyInline(admin.TabularInline):
model = models.mymodel
can_delete = False
I found a very easy solution to quietly avoid unwanted deletion of some inlines. You can just override delete_forms property method.
This works not just on admin, but on regular inlines too.
from django.forms.models import BaseInlineFormSet
class MyInlineFormSet(BaseInlineFormSet):
#property
def deleted_forms(self):
deleted_forms = super(MyInlineFormSet, self).deleted_forms
for i, form in enumerate(deleted_forms):
# Use form.instance to access object instance if needed
if some_criteria_to_prevent_deletion:
deleted_forms.pop(i)
return deleted_forms
I am using Django 1.3 and python 2.7 .I am using Django admin app.What I want is when a superuser logs-in it should be shown admin/index.html with all models which is default behaviour but if any other user logs-in that is not superuser then it should be shown a totally different template with my data (like 'abc.html').What should I do to accomplish this?I guess I need to override some admin view to do this but have no idea?
Please help.If you want more information plz comment :)
I would create a middleware that checks if the user is a superuser or not. If the user is not supeuser you redirects him/her to the custom admin page instead of the default one.
class SuperUserMiddleware(object):
def process_request(self, request):
user = request.session.user
if not user.is_superuser:
return HttpResponseRedirect(NON_SUPERUSER_URL)
...
You create a modified AdminSite class definition with additional permission rules.
class SuperUserAdminSite( AdminSite ):
def has_permission(self, request):
return request.user.is_active and request.user.is_staff and request.user. is_superuser
Now you can create two AdminSite objects, one for ordinary users, one for super users.
You can have two paths in your URLs for the two admin sites.
Superusers can use both paths.
Ordinary users will only be able to use the ordinary user path in the URL.
https://docs.djangoproject.com/en/1.3/ref/contrib/admin/#adminsite-objects
You have to change the view of the admin site. Django Documentation mention all in detail. Please check that https://docs.djangoproject.com/en/1.3/ref/contrib/admin/ if you have any error then please write back with some code details.