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Allowing only the specific users or group of users to visit the specific page of the website in Django?

Let us suppose that there is a admin page in my website designed by me for doing some actions on the website.
now i want that only few user that belongs to a specific can visit that page.
other users should redirected or should have see an error page when they try to go at that page.
for
karan(admin ) can visit that page
but not ram(simple user)
you can also refer any link from the django docs
thanks!

You can use the permission required decorator for function based views or the permission required mixin for class based views.
https://docs.djangoproject.com/en/3.2/topics/auth/default/#the-permission-required-decorator
To create a custom permission you can use the PermissionsMixin on your models.
class CustomUser(AbstractBaseUser, PermissionsMixin):
class Meta:
permissions = (
("validator","Can access validation views"),
("administrator","Almost all permissions, but not a superuser"),
("teacher","user to create question"),
)

Django - How to differentiate between users in the database?

I'm building a Django server for my company and I'm still unfamiliar with some processes. I'm sure this is super simple, I'm just completely unaware of how this works.
How do I differentiate between user's data so it doesn't get mixed up?
If Jill is a user and she requests a page of her profile data, how do I not send her Jack's profile data, especially if there are multiple models invovled?
For example, the code in the view would look like this:
def display_profile(request)
profile = Profile.objects.get(???) # What do I put in here?
I understand that I can do:
def display_profile(request, user)
profile = Profile.objects.get(user_id=user)
But that's not my design intention.
Thank you in advance.

As documented
Django uses sessions and middleware to hook the authentication system into request objects.
These provide a request.user attribute on every request which
represents the current user. If the current user has not logged in,
this attribute will be set to an instance of AnonymousUser, otherwise
it will be an instance of User.
So in your case (notice field not being called user_id )
profile = Profile.objects.get(user=user)

In your Django view, you can access the current user with request.user.
So if you want to get a Profile instance matching your current logged in user, just do a query as follow:
profile = Profile.objects.get(user=request.user)
This assumes you have a user foreign key field (or OneToOne) in your Profile model.

Wagtail Admin: How to control where user is redirected after submitting changes for moderation

On a Wagtail site I've built, I have a model type that is editable by authenticated users who do not have full admin privileges. They can only save as a draft or submit the changes for moderation. The problem I'm having is that Wagtail is inconsistent about where it redirects after executing these two actions. Saving the Draft takes the user back to the edit screen they were just on, with a note saying the draft was saved (good). Submitting for Moderation returns the user to the admin browse view of the parent page, which shows all of the sibling nodes in a list. They can't edit the vast majority of items on that list, so I think this is confusing for a non-admin user. I would like to have the "Submit for Moderation" action detect whether the user belongs to a group other than admin (or, failing that, whether the page has unpublished changes, as in my code example below) and, if so, redirect them back to the edit screen just like "Save as Draft" does.
I tried this in my model definition and it didn't work:
def save(self, *args, **kwargs):
#do some field value manipulations here before saving
super().save(*args, **kwargs)
if self.id:
if self.has_unpublished_changes:
return HttpResponseRedirect('/admin/pages/' + str(self.id) + '/edit/')
There's probably some sort of Wagtail admin action that I need to hook into and override, rather than trying to accomplish this in models.py, but I don't have much experience with this, so I need a better understanding of what to change and where.

Set up an after_create_page and after_edit_page hook: https://docs.wagtail.io/en/stable/reference/hooks.html#after-create-page
To do this, add a wagtail_hooks.py file in one of the apps in your project. This is where you can define functions to be called after creating or editing a page through the admin, such as:
from wagtail.core import hooks
#hooks.register('after_create_page')
def redirect_after_page_create(request, page):
if not request.user.is_superuser:
return HttpResponseRedirect('/admin/pages/' + str(page.id) + '/edit/')
#hooks.register('after_edit_page')
def redirect_after_page_edit(request, page):
if not request.user.is_superuser:
return HttpResponseRedirect('/admin/pages/' + str(page.id) + '/edit/')

How do I separate user accounts in Django ?

I am using Django to create an app that allows recording of medical information however I am having problems with seperating the user accounts so currently all users see the same information entered. Anyone familiar with django knows how to set the proper permissions and roles and is willing to help a newby out?
I want the user to only access to the account the user creates and the records that the user create.
This is my github link
If you are able to to help I would really appreciate it.

If you want to list only the user's records in your /home . You only need to change the query in your home/views.py, from Identity_unique.objects.all() to Identity_unique.objects.filter(user=request.user)
class Identity_view(TemplateView):
def get(self, request):
form = Identity_form()
Identities = Identity_unique.objects.filter(user=request.user)
var = {'form': form, 'Identities': Identities}
return render(request, self.template_name, var)
Or if you want to filter objects in your Django Admin panel you should read this:
Django Documentation: ModelAdmin.get_queryset(request)

Create a custom user model with an extra field user_type.
https://github.com/samimsk/debatehub/blob/master/devdebatehub/UserApp/models.py
Implemented here.

Multiple Instances of a django app, does django support this

I have written a simple feedback application in django. It's not particulairly complex, basically it allows authenticated users to write a shot message with a subject line and submit that message via a form. I then allows who are in a selected group to view user submitted feedback. In the future I may add more functionality but for now it does what I want.
Here comes my question, the site I'm building has multiple places where I would like to use the feedback app, for example I have a "what do you think of the site?" kind of page at /dev/feedback/ I also have one for customer support feedback at "/support/feedback/" Currently I have just copied the code from my mysite.apps.dev.feedback over to mysite.apps.support.feedback.
The problem is that this has now created two separate copies of the same code. Despite having just written the app the two versions are already starting to diverge which is annoying. My question is simply how do I create multiple instances of the same app in a django site with distinct database models?
Some resources I've found related but not helpful are https://docs.djangoproject.com/en/dev/topics/http/urls/ and Reversing namespaced URLs in Django: multiple instances of the same app The first page does not offer much on the issue and the second page provides somewhat cludgey and impractical solutions that seem to be both unrelated and more work than their worth. Is there a proper way to implement multiple instances of the same django app?

Single model approach
I'd personally try to keep this as one app and have a view that can handle being posted from multiple locations / tag them appropriately.
As S.Lott says, this is the way to go. I am providing alternatives if you're curious about methods to keep your code in one place in other situations.
For example, you could add a category field to your model, set up a single url conf which accepts an argument in the URL such as /(?P<category>\w+/feedback/$ and have the view simply tag the feedback with the appropriate category.
class MyForm(forms.ModelForm):
class Meta:
model = Feedback
def my_view(request, category):
form = MyForm(request.POST or None)
if request.method == 'POST':
if form.is_valid():
feedback = form.save(commit=False)
feedback.category = category
feedback.save()
return http.HttpResponse("Thanks for posting!")
return render(request, "mytemplate.html", {'form': form})
# urls.py
(r'^(?P<category>\w+)/feedback/$', 'my_view')
# user can visit dev/feedback or support/feedback and the feedback will be tagged appropriately
Abstract base class
Another solution is to build an abstract base class, then create subclasses for your distinct tables. That should solve the issue with your code getting out of sync.
You'd have a single abstract model (which has no tables) from which your "real" models in your separate apps would be based on.
Dynamically generated views
If you must have separate models, you could potentially write a dynamically constructed view.
def view_generator(model_class):
class MyForm(forms.ModelForm):
class Meta:
model = model_class
def my_view(request):
form = MyForm(request.POST or None)
if request.method == 'POST':
if form.is_valid():
form.save()
return http.HttpResponse("Thanks for posting!")
return render(request, "mytemplate.html", {'form': form})
return my_view
# urls.py
from foo import view_generator
(r'^my_first_feedback_form', view_generator(Model1))
(r'^my_second_feedback_form', view_generator(Model2l))

how do I create multiple instances of the same app in a django site with distinct database models?
You shouldn't.
You simply use the feedback app model in the other two apps with a simple from feedback.models import Feedback.
Then your support app can create, retrieve, update and delete Feedback objects.
Your dev app, similarly, can create, retrieve, update and delete Feedback objects because it imported the model.
That's all that's required: import.

Thanks Yuji Tomita for a very thorough answer, my final solution is derived very closely from his suggestion, but is different enough that I thought I would post it as another option if someone else runs into the same situation that I am in.
Firstly in my mysite.apps.feedback.models file I put
class Feedback( models.Model ):
subject = models.TextField( max_length=100 )
body = models.TextField( max_length=100 )
# Some other stuff here...
# Finally I used the suggestion above and created a field which I
# use to label each entry as belonging to a specific instance of the app.
instance_name = models.TextField( max_length=20 )
In my mysite.apps.feedback.views file I put
def save_message( request, instance_name ):
if request.method == 'POST':
form = FeedbackFrom( request.POST )
if form.is_valid():
form.instance.instance_name = instance_name
form.save()
return render("feedback/thanks.html")
else:
return render("feedback/submit.html", {'form':form })
else:
return render("feedback/submit.html",{'form':FeedbackForm()})
#user_passes_test( is_staff )
def all_messages( request, instance_name ):
messages = Feedback.objects.filter( instance_name = instance_name )
return render("feedback/view_all.html",{'feedback':messages} )
In my mysite.apps.dev.urls file I put
url(r'^feedback/', include('mysite.apps.feedback.urls'),
{'instance_name':'dev'}),
In my mysite.apps.support.urls file I put
url(r'^feedback/', include('mysite.apps.feedback.urls'),
{'instance_name':'support'}),
This will separate feedback messages by app instance. Note that my actual code is more complex but this should be good enough for anyone with a similar problem to get a solution up and running pretty quickly. Hope this is useful to anyone in a similar situation. Thanks again to Yuji Tomita for the suggestions upon which this solution is based.