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How can I redirect in django middleware? global name 'view' is not defined

Django newbie here, need help on basic middleware to redirect to another view if a certain model field is empty.
I am creating a terms of agreement page that users must get redirected to right after they signup to the platform if their filed_terms field on their Profile model is empty.
I am using middleware for this. However I am unable to get this to work. This is my middleware class:
class TermsMiddleware(object):
def process_request(self, request):
if request.user.profile.filled_terms is None:
return redirect(reverse(terms))
This gives me the following error:
global name 'terms' is not defined
I also have the url matcher that works perfectly when I navigate to it manually:
url(r'^terms/', 'my_app.views.terms')
I have a terms.html template and a terms view in my views.py file that is working perfectly in all other respects. I have also added it to the settings middleware requirements to make sure it loads.
Do I have to import something from views or url dispatcher into my middleware file? If so what would that be? I have been at this for a while an cannot find anything helpful.

reverse function takes url name instead on the regex. So you need to add name on your url configuration. Here is the example.
url(r'^terms/', 'my_app.views.terms', name='terms')
Add this in your views.py
from django.core.urlresolvers import reverse
And you need to fix your reverse function into.
return redirect(reverse('terms'))
Python interpret your terms as a variable and you have no variable named terms while you need to put string on reverse.

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)

How to reference a ModelView in flask-admin

What's the right way to get the URL for a flask-admin ModelView?
Here's a very simple example:
my_admin_view.py
from flask.ext.admin.contrib.sqla import ModelView
from common.flask_app import app
from models import db, User, Role
admin = Admin(app, name="Boost Admin")
admin.add_view(ModelView(User, db.session, category="model"))
admin.add_view(ModelView(Role, db.session, category="model"))
my_admin_template.html
...
<p>Check out my user admin link:</p>
User view link
{# ______________
what argument to pass in here?? #}
...
What's the correct argument to pass to url_for(...)?
I've tried modelview.user, my_admin_view.modelview.user, etc. None of them seem to resolve correctly, and I'd like to avoid hardcoding the link.
thanks!

OK I figured it out after reading the source code for ModelView.
First, make sure that endpoints are named (it's possible to do it without named endpoints, but this makes the code much clearer):
from flask.ext.admin.contrib.sqla import ModelView
from models import db, User, Role
admin = Admin(app, name="Boost Admin")
admin.add_view(ModelView(User,db.session,category="model", endpoint="model_view_user"))
admin.add_view(ModelView(Role,db.session,category="model", endpoint="model_view_role"))
...now in the template, you can reference the basic model view as follows:
URL for User model default view is: {{model_view_user.index_view}}
URL for Role model default view is: {{model_view_role.index_view}}
The index_view function is defined here, and implements the default view for a flask admin ModelView.

See the section Generating URLs in the Flask-Admin introduction.
It says to "use the lowercase name of the model as the prefix". Add a dot, and the name of the view.
index_view for the overview list.
create_view for creating a new row.
edit_view for modifying an existing row.
So you can easily do:
url_for('user.index_view')
url_for('role.create_view')
url_for('user.edit_view', id=1)

It should be
url_for('admin.user')
If you read the flask-admin docs here, for generating URLs, it clearly says:
If you want to generate a URL for a particular view method from outside, the following rules apply:
....
3. For model-based views the rules differ - the model class name should be used if an endpoint name is not provided.

How to use current logged in user as PK for Django DetailView?

When defining URL patterns, I am supposed to use a regular expression to acquire a PK from the URL.
What if I want a URL that has no PK, and if it's not provided, it will use the currently logged in user? Examples:
visiting /user will get a DetailView of the currently logged in user
/user/edit will show an UpdateView for the currently logged in user
I tried hard-coding the pk= in the Detail.as_view() call but it reports invalid keyword.
How do I specify that in the URL conf?
My sample code that shows PK required error when visiting /user URL:
urlpatterns = patterns('',
url(r'user/$',
DetailView.as_view(
model=Account,
template_name='user/detail.html')),
)`

An alternative approach would be overriding the get_object method of the DetailView subclass, something along the line of:
class CurrentUserDetailView(UserDetailView):
def get_object(self):
return self.request.user
Much cleaner, simpler and more in the spirit of the class-based views than the mixin approach.
EDIT: To clarify, I believe that two different URL patterns (i.e. one with a pk and the other without) should be defined separately in the urlconf. Therefore they could be served by two different views as well, especially as this makes the code cleaner. In this case the urlconf might look something like:
urlpatterns = patterns('',
url(r"^users/(?P<pk>\d+)/$", UserDetailView.as_view(), name="user_detail"),
url(r"^users/current/$", CurrentUserDetailView.as_view(), name="current_user_detail"),
url(r"^users/$", UserListView.as_view(), name="user_list"),
)
And I've updated my example above to note that it inherits the UserDetailView, which makes it even cleaner, and makes it clear what it really is: a special case of the parent view.

As far as I know, you can't define that on the URL definition, since you don't have access to that information.
However, what you can do is create your own mixin and use it to build views that behave like you want.
Your mixin would look something like this:
class CurrentUserMixin(object):
model = Account
def get_object(self, *args, **kwargs):
try:
obj = super(CurrentUserMixin, self).get_object(*args, **kwargs)
except AttributeError:
# SingleObjectMixin throws an AttributeError when no pk or slug
# is present on the url. In those cases, we use the current user
obj = self.request.user.account
return obj
and then, make your custom views:
class UserDetailView(CurrentUserMixin, DetailView):
pass
class UserUpdateView(CurrentUserMixin, UpdateView):
pass

Generic views uses always RequestContext. And this paragraph in the Django Documentation says that when using RequestContext with auth app, the template gets passed an user variable that represents current user logged in. So, go ahead, and feel free to reference user in your templates.

You can get the details of the current user from the request object. If you'd like to see a different user's details, you can pass the url as parameter. The url would be encoded like:
url(r'user/(?P<user_id>.*)$', 'views.user_details', name='user-details'),
views.user_details 2nd parameter would be user_id which is a string (you can change the regex in the url to restrict integer values, but the parameter would still of type string). Here's a list of other examples for url patterns from the Django documentation.

Overriding admin views - Django

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.