Im following django tutorial and there is additional ideas how to test my code.
Perhaps logged-in admin users should be allowed to see unpublished
Questions, but not ordinary visitors. Again: whatever needs to be
added to the software to accomplish this should be accompanied by a
test,...
How can I create test to check if logged in User can see Questions witout choices and not logged in not?
class QuestionsAndChoices(TestCase):
def test_user_can_see_question_without_choices(self):
"""
A Question without choices should be displayed only for logged in Users
"""
#first make an empty question to use as a test
empty_question = create_question(question_text='Empty question', days=-1)
#so we can create choice but in this case we don't need it
#answer_for_question = past_question.choice_set.create(choice_text='Answer for "Question"', votes=0)
#create a response object to simulate someone using the site
response = self.client.get(reverse('polls:index'))
#if user logged in output should contain the question without choices
self.assertQuerysetEqual(response.context['latest_question_list'], []) #returns empty querylist
In the setUp method of your test class you can create a user.
class QuestionsAndChoices(TestCase):
def setUp(self):
self.user = User.objects.create_user(
username='user', password='top_secret')
Then in your test method, you can use force_login to log the user in, then use self.client.get() and make assertions as usual.
class QuestionsAndChoices(TestCase):
def test_user_can_see_question_without_choices(self):
...
self.client.force_login(self.user)
#create a response object to simulate someone using the site
response = self.client.get(reverse('polls:index'))
#if user logged in output should contain the question without choices
...
Related
I have created unique user pages for each person who registers to this social app by using the following:
#app.route("/profile/<username>", methods=["GET", "POST"])
def profile(username):
user = session["user"] or None
# grab the session user's username from db
username = mongo.db.users.find_one(
{"username": session["user"]})["username"]
user_profile = mongo.db.users.find_one({"username": user})
if session["user"]:
return render_template("profile.html", username=username, profile=user_profile)
return redirect(url_for("login"))
This creates a unique username profile built up from edited data that the user puts in. But I want to be able to view that profile from another user's login and then add them as a friend (if I choose).
Because I already have a URL created /profile/ do I need to create a new #app.route to something such as view_profile/<username>?
From there, I would use the Post and Get methods so if someone wanted to add that user as a friend they could click on that. From there, I think I would need the other user to accept the friend request? Not quite sure how to get to that just yet. But one step at a time I guess.
I started with the idea of a view_profile #app.route but now I am not 100% sure which direction to take it that will make the next stages also easy.
This is more a process logic question than a specific language-framework one.
I am developing a mobile app and want the user to be able to use it without having to login (i.e. try it and offer a plus to the logged users), but I don´t want other persons to make post requests from let´s say Postman or any other platform than the app without having some sort of key, so what would be the approach here?
I am thinking on basic auth with some secret username:password for guests, or some kind of token, but as I am totally new on this I am not sure if it´s the correct approach, I´ve read the authentication and permissions Django Rest Framework tutorial but haven´t found a solution
I am learning Django myself and have gotten to the more advanced topics in the subject. What you could do is create a function in your permissions.py file for this. like so:
from rest_framework import permissions
class specialMobileUserPermissions(permissions.BasePermission):
def has_object_permission(self, request, view, obj):
if request.method in request.SAFE_METHODS:
return True
if request.user.id == whatever your mobile users id is:
return false
return obj.id == request.user.id # if the user is a subscribed user and they are logged in return true
return false # because we need a way out if none of the above works
So when dealing with permissions classes the permissions.SAFE_PERMISSIONS is a list of permissions that are non-destructive. So the first if statement asks are you a GET, HEAD, or other non data altering method. If so return true.
The second if statement checks the user id of the user that is making the request. And if that user id is equal to the user id you set for the mobile trail user it would return false, denying permissions to whatever this class is used on.
In your viewset you would need to add the permissions_classes variable like below
from . import permissions # your permissions.py file
class FooViewSet(viewsets.ViewSet):
permission_classes = (permissions.specialMobileUserPermissions,)
Unless you need extra functionality, that should be everything you need, all the way down to the imports. I hope I have helped.
I have setup a basic Django site and have added login to the site. Additionally, I have created a Student (Profile) model that expands upon the built in User one. It has a OneToOne relationship with the User Model.
However, I have yet not come right with forcing the user to automatically create a Profile the first time they log in. How would I make sure they are not able to progress through anything without creating this?
I have tried by defining the following in views:
def UserCheck(request):
current_user = request.user
# Check for or create a Student; set default account type here
try:
profile = Student.objects.get(user = request.user)
if profile == None:
return redirect('/student/profile/update')
return True
except:
return redirect('/student/profile/update')
And thereafter adding the following:
UserCheck(request)
at the top of each of my views. This however, never seems to redirect a user to create a profile.
Is there a best way to ensure that the User is forced to create a profile object above?
Looks like you're attempting to do something similar to Django's user_passes_test decorator (documentation). You can turn the function you have into this:
# Side note: Classes are CamelCase, not functions
def user_check(user):
# Simpler way of seeing if the profile exists
profile_exists = Student.objects.filter(user=user).exists()
if profile_exists:
# The user can continue
return True
else:
# If they don't, they need to be sent elsewhere
return False
Then, you can add a decorator to your views:
from django.contrib.auth.decorators import user_passes_test
# Login URL is where they will be sent if user_check returns False
#user_passes_test(user_check, login_url='/student/profile/update')
def some_view(request):
# Do stuff here
pass
I want to make users active by sending them an activation email to click. I guess it is currently not incorporated in Django 1.6.The user-registration app coded in Django seems to serve this purpose. But I have some doubts with regard to the DefaultForm it provides in forms.py. I want to have more fields included in it. How can I achieve that in class RegistrationForm(forms.Form) implemented there. If I install this app, is it a good idea to change include more fields directly there, is there a better way to achieve the same.
In the views.py, I see some methods such as the following are not implemented. I dont have a clear picture of what these methods need to do. should I redirect the url here to the pages?
def register(self, request, **cleaned_data):
raise NotImplementedError
def activate(self, request, *args, **kwargs):
raise NotImplementedError
def get_success_url(self, request, user):
raise NotImplementedError
You need to first let them sign up and mark them as is_active=False for the time being. Something like this:
from django.contrib.auth.models import User
from django.core.mail import send_mail
from django.http import HttpResponseRedirect
def signup(request):
# form to sign up is valid
user = User.objects.create_user('username', 'email', 'password')
user.is_active=False
user.save()
# now send them an email with a link in order to activate their user account
# you can also use an html django email template to send the email instead
# if you want
send_mail('subject', 'msg [include activation link to View here to activate account]', 'from_email', ['to_email'], fail_silently=False)
return HttpResponseRedirect('register_success_view')
Then once they click the link in the email it takes them to the next view (note: you need to put a link in the email so that you know which user it is. This may be 16-digit salt or something. The below view uses the user.pk:
def activate_view(request, pk):
user = User.objects.get(pk=pk)
user.is_active=True
user.save()
return HttpResponseRedirect('activation_success_view')
Hope that helps. Good Luck!
Basically you can use django's user model(https://docs.djangoproject.com/en/1.9/ref/contrib/auth/). However, in user model email is not a required field. You need to modify the model to make email an required field.
In your views, you might need the following method:
1) Sign up: after sign up, set user.is_active=False and call function send_email to include an activation link in the email. In the link, you might want to include the user's information (for example, user.id) so when the user click the link, you know which user to activate.
2) send_email: send a link to user's email address for verification. The link includes user's id. For example:
http://127.0.0.1:8000/activation/?id=4
3) Activate: get the id information from the URL using id=request.GET.get('id'). Query user=user whose id is id. set user.is_active=True.
Actually I implemented an reusable application like your request, check this(https://github.com/JunyiJ/django-register-activate) if you are interested.
Hope that helps. Good luck!
check this out... i Hope it helps out with not only the solution u need but also the explanation.. Because i think django-registration app is meant for default Django User. So if u want to have extra fields in your registration form, Start thinking of customizing ur Django User and Its authentication yourself. You dont need the django-registration app here..
Here are some tutorials thats will help
http://www.caktusgroup.com/blog/2013/08/07/migrating-custom-user-model-django/
and many more...
I am relatively new with Django and it's ecosystem. I am writing REST api for our mobile client using django-tastypie. I have gone through almost all the examples on the web about how to use tastypie for creating REST interfaces. but none of them are specific to POSTing the data from client and how would you authorize a client.
I used the from tastypie.authentication.BasicAuthentication as show in the example. It opens a pop up asking username and password and works fine on the browser. But I am not sure, if it will do the same thing on mobile (to be specific, native IOS app). I am not quite getting when a user will make a request to login how this popup will be shown there on his/her mobile device if he or she is not using the browser but the native app.
I am totally lost on this, I would really appreciate your help.
You can check out source and use for example ApiKeyAuthentication.
You just have to POST username and api key to authentificate user.
It looks like usable for ios app.
Here is the part of the checking code.
def is_authenticated(self, request, **kwargs):
"""
Finds the user and checks their API key.
Should return either ``True`` if allowed, ``False`` if not or an
``HttpResponse`` if you need something custom.
"""
from django.contrib.auth.models import User
username = request.GET.get('username') or request.POST.get('username')
api_key = request.GET.get('api_key') or request.POST.get('api_key')
if not username or not api_key:
return self._unauthorized()
try:
user = User.objects.get(username=username)
except (User.DoesNotExist, User.MultipleObjectsReturned):
return self._unauthorized()
request.user = user
return self.get_key(user, api_key)
https://github.com/toastdriven/django-tastypie/blob/master/tastypie/authentication.py#L128
https://github.com/toastdriven/django-tastypie/blob/master/tastypie/authorization.py#L42
Thanks for the help.
I used similar approach mentioned by #Iurii. Here is my solution.
I wrote a class for handling the authentication and override is_authenticated method. and then I can use this class in Meta definition of tastypie resource classes.
from tastypie.authentication import BasicAuthentication
from tastypie.resources import Resource, ModelResource
# class for handling authentication
class MyAuthentication(BasicAuthentication):
def is_authenticated(self, request, **kwargs):
# put here the logic to check username and password from request object
# if the user is authenticated then return True otherwise return False
# tastypie resource class
class MyResource(ModelResource):
class Meta:
authentication = MyAuthentication()
this will ensure a request to access the resource will go through your authentication code.