I have several function that need to have a 'redirect' filter. The redirect filter goes something like this --
1) if a user is not logged in and has no session data, redirect to login page.
2) if a user is logged in and has already filled out the page, redirect to user home.
3) if a user is logged in and has not already filled out the page, stay on the page.
4) if a user is not logged in and has session data, stay on the page
I've started to convert the functions into a class-based approach to make it more efficient and less code (previously my view functions were pretty massive). This is my first stab at trying make something class-based, and this is what I have so far --
def redirect_filter(request):
if request.user.is_authenticated():
user = User.objects.get(email=request.user.username)
if user.get_profile().getting_started_boolean:
return redirect('/home/') ## redirect to home if a logged-in user with profile filled out
else:
pass ## otherwise, stay on the current page
else
username = request.session.get('username')
if not username: ## if not logged in, no session info, redirect to user login
return redirect('/account/login')
else:
pass ## otherwise, stay on the current page
def getting_started_info(request, positions=[]):
location = request.session.get('location')
redirect_filter(request)
if request.method == 'POST':
form = GettingStartedForm(request.POST)
...(run the function)...
else:
form = GettingStartedForm() # inital = {'location': location}
return render_to_response('registration/getting_started_info1.html', {'form':form, 'positions': positions,}, context_instance=RequestContext(request))
Obviously, this view is not fully working yet. How would I convert this into something that's functional?
Also, I have three variables that will need to be reused in several of the getting_started functions:
user = User.objects.get(email=request.user.username)
profile = UserProfile.objects.get(user=user)
location = profile.location
Where would I put these variable definitions so I can reuse them in all the functions, and how would I call them?
Thank you.
Django actually already includes a login_required decorator that makes handling user authentication trivial. Just include the following at the top of your view.py page:
from django.contrib.auth.decorators import login_required
and then add
#login_required
before any views that require a login. It even handles redirecting the user to the appropriate page once they log in.
More info here:
https://docs.djangoproject.com/en/dev/topics/auth/#the-login-required-decorator
This should greatly simplify your views, and may result in not having to write a separate class, since all that's left is a simple re-direct.
As for the variables, each request already contains a request.user object with information on the user. You can do a search in the docs for Request and response objects to learn more.
You can use that user object to get the profile variable by extending the user module. Set AUTH_PROFILE_MODULE = 'myapp.UserProfile' in your Settings, which will allow you to access a users profile as follows:
user.get_profile().location.
More about that here:
http://www.b-list.org/weblog/2006/jun/06/django-tips-extending-user-model/
Related
Okay so, ordinary Django allows you to simply:
if request.user.is_authenticated:
I want to be able to do the same in Pyrebase. Have the views sort of already know which user has logged in based on the current session without having to sign the user in in all views.
I have tried:
def sign_in(request):
user = firebase.auth().sign_in_with_email_and_password('email', 'password')
user_token = firebase.auth().refresh(user['refreshToken']
request.session['session_id'] = user_token
I noticed this creates a session ID for me. But I don't know how to associate it with the current user and I know it has something to do with the refresh token.
If I don't check authentication, anyone can visit any page of my site without signing in.
I am working through the Pyramid authorization tutorial and I have noticed the pattern where
logged_in = request.authenticated_userid
is added to each view dictionary. Can it be avoided? I.e. is there a configuration which automatically ads user id to each view. Or is there a way to create a base, abstract view with the user id and inherit from it?
Part of the code from the tutorial:
#view_config(context='.models.Page', renderer='templates/view.pt', permission='view')
def view_page(context, request):
# not relevant code
return dict(page = context, content = content, edit_url = edit_url,
logged_in = request.authenticated_userid)
#view_config(name='add_page', context='.models.Wiki', renderer='templates/edit.pt',
permission='edit')
def add_page(context, request):
# not relevant code
return dict(page=page, save_url=save_url,
logged_in=request.authenticated_userid)
It's been awhile since I last looked, but I think logged_in in the samples is just an example to use to conditionally check if there is a logged on user or not. You could probably just as easily refer to request.authenticated_userid within any of your views or templates, too, and get the same behavior and not have to explicitly add a status to the response dict. The request object should be available to be referenced in your view templates, too.
Alternatively, I've used their cookbook to add a user object to the request to make a friendly request.user object that I can use to both check for logged in status where needed, plus get at my other user object details if I need to as well.
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)
Question Clarification:
I'm trying to test if the user is authenticated or not for each page request.
I'm trying to use Authentication for the first time in Django and I am not grasping how the login view is supposed to handle authentications.
When I use #login_required, I'm redirecting to "/login" to check if the user is logged in and if not, display the login page. However, trying to redirect back to the original page is causing an infinite loop because it's sending me back to the login page over and over again.
I'm clearly not grasping how #login_required is supposed to work but I'm not sure what I'm missing. I've been searching around for awhile for an example, but everyone uses the default #login_required without the 'login_url' parameter.
So for example.. the page I'm trying to access would be...
#login_required(login_url='/login')
def index(request):
And then my login would be.. (obviously incomplete)..
Edit: just to note.. the session variables are set in another view
def login(request):
if '_auth_user_id' in request.session:
# just for testing purposes.. to make sure the id is being set
print "id:",request.session['_auth_user_id']
try:
user = Users.objects.get(id=request.session['_auth_user_id'])
except:
raise Exception("Invalid UserID")
# TODO: Check the backend session key
# this is what I'm having trouble with.. I'm not sure how
# to redirect back to the proper view
return redirect('/')
else:
form = LoginForm()
return render_to_response('login.html',
{'form':form},
context_instance=RequestContext(request)
)
Well, as you say, obviously that's not going to work, because it's incomplete. So, until you complete it, you're going to get an infinite loop - because you haven't written the code that puts _auth_user_id into request.session.
But I don't really know why you're making that test in the first place. The auth documentation has a perfectly good example of how to write a login view: get the username and password from the form, send them to authenticate to get the user object, then pass that to login... done.
Edit I think I might see where your confusion is. The login_required decorator itself does the check for whether the user is logged in - that's exactly what it's for. There's no need for you to write any code to do that. Your job is to write the code that actually logs the user in, by calling authenticate and login.
Try to call login(), see the next please:
https://docs.djangoproject.com/en/dev/topics/auth/#django.contrib.auth.login
I have a Django app. When logged in as an admin user, I want to be able to pass a secret parameter in the URL and have the whole site behave as if I were another user.
Let's say I have the URL /my-profile/ which shows the currently logged in user's profile. I want to be able to do something like /my-profile/?__user_id=123 and have the underlying view believe that I am actually the user with ID 123 (thus render that user's profile).
Why do I want that?
Simply because it's much easier to reproduce certain bugs that only appear in a single user's account.
My questions:
What would be the easiest way to implement something like this?
Is there any security concern I should have in mind when doing this? Note that I (obviously) only want to have this feature for admin users, and our admin users have full access to the source code, database, etc. anyway, so it's not really a "backdoor"; it just makes it easier to access a user's account.
I don't have enough reputation to edit or reply yet (I think), but I found that although ionaut's solution worked in simple cases, a more robust solution for me was to use a session variable. That way, even AJAX requests are served correctly without modifying the request URL to include a GET impersonation parameter.
class ImpersonateMiddleware(object):
def process_request(self, request):
if request.user.is_superuser and "__impersonate" in request.GET:
request.session['impersonate_id'] = int(request.GET["__impersonate"])
elif "__unimpersonate" in request.GET:
del request.session['impersonate_id']
if request.user.is_superuser and 'impersonate_id' in request.session:
request.user = User.objects.get(id=request.session['impersonate_id'])
Usage:
log in: http://localhost/?__impersonate=[USERID]
log out (back to admin): http://localhost/?__unimpersonate=True
It looks like quite a few other people have had this problem and have written re-usable apps to do this and at least some are listed on the django packages page for user switching. The most active at time of writing appear to be:
django-hijack puts a "hijack" button in the user list in the admin, along with a bit at the top of page for while you've hijacked an account.
impostor means you can login with username "me as other" and your own password
django-impersonate sets up URLs to start impersonating a user, stop, search etc
I solved this with a simple middleware. It also handles redirects (that is, the GET parameter is preserved during a redirect). Here it is:
class ImpersonateMiddleware(object):
def process_request(self, request):
if request.user.is_superuser and "__impersonate" in request.GET:
request.user = models.User.objects.get(id=int(request.GET["__impersonate"]))
def process_response(self, request, response):
if request.user.is_superuser and "__impersonate" in request.GET:
if isinstance(response, http.HttpResponseRedirect):
location = response["Location"]
if "?" in location:
location += "&"
else:
location += "?"
location += "__impersonate=%s" % request.GET["__impersonate"]
response["Location"] = location
return response
#Charles Offenbacher's answer is great for impersonating users who are not being authenticated via tokens. However, it will not work with clients side apps that use token authentication. To get user impersonation to work with apps using tokens, one has to directly set the HTTP_AUTHORIZATION header in the Impersonate Middleware. My answer basically plagiarizes Charles's answer and adds lines for manually setting said header.
class ImpersonateMiddleware(object):
def process_request(self, request):
if request.user.is_superuser and "__impersonate" in request.GET:
request.session['impersonate_id'] = int(request.GET["__impersonate"])
elif "__unimpersonate" in request.GET:
del request.session['impersonate_id']
if request.user.is_superuser and 'impersonate_id' in request.session:
request.user = User.objects.get(id=request.session['impersonate_id'])
# retrieve user's token
token = Token.objects.get(user=request.user)
# manually set authorization header to user's token as it will be set to that of the admin's (assuming the admin has one, of course)
request.META['HTTP_AUTHORIZATION'] = 'Token {0}'.format(token.key)
i don't see how that is a security hole any more than using su - someuser as root on a a unix machine. root or an django-admin with root/admin access to the database can fake anything if he/she wants to. the risk is only in the django-admin account being cracked at which point the cracker could hide tracks by becoming another user and then faking actions as the user.
yes, it may be called a backdoor, but as ibz says, admins have access to the database anyways. being able to make changes to the database in that light is also a backdoor.
Set up so you have two different host names to the same server. If you are doing it locally, you can connect with 127.0.0.1, or localhost, for example. Your browser will see this as three different sites, and you can be logged in with different users. The same works for your site.
So in addition to www.mysite.com you can set up test.mysite.com, and log in with the user there. I often set up sites (with Plone) so I have both www.mysite.com and admin.mysite.com, and only allow access to the admin pages from there, meaning I can log in to the normal site with the username that has the problems.