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Why not all showing up in vs code django/python intellisense

Hi I am new to Python/django and am using VS Code. Now I got python IntelliSense and pylance extension installed and most things are showing up on IntelliSense but some aren't.
How can I get it to work for everything?
I would very much appreciate some insight since it's driving me nuts...
request.POST from an imported lib not showing up
selected_meetup.participants not showing up nor is selected_meetup.participants.add
from urllib import request
from django.forms import SlugField
from django.shortcuts import render
from .models import Meetup
from .forms import RegistrationForm
# from django.http import HttpResponse
# Create your views here.
def index(request):
meetups = Meetup.objects.all()
return render(request, 'meetups/index.html', {
'meetups': meetups
})
def meetup_details(request, meetup_slug):
try:
selected_meetup = Meetup.objects.get(slug=meetup_slug)
if request == 'GET':
registration_form = RegistrationForm()
else:
registration_form = RegistrationForm(request.POST)
registration_form = RegistrationForm(request.)
if registration_form.is_valid():
participant = registration_form.save()
selected_meetup.participants.add(participant)
return render(request, 'meetups/meetup-detail.html', {
'meetup_found': True,
'meetup': selected_meetup,
'form': registration_form
})
except Exception as exc:
return render(request, 'meetups/meetup-detail.html', {'meetup_found': False
})
Update:
That's great guys, I now understand the type problem it's obvious now (only had so many extensions issues especially django/html) I got confused...
Now I added types to Meetup and Participant but selected_meetup.participants. still refused to show me its method ADD.
I assume the problem is the lib is written without types. Is there a way to solve that? because quite valuable to me to be able to see what's possible write away...
have you got solutions to that?

I would re-install the extensions, then make sure it is referencing the right version of python. If that doesn't work I would recommend to just type it out in full and look past the error.

You didn't declare what type of object request is in the programming process. Maybe you can declare request = request () in the code.The second is the same. vscode can't recognize the type of the selected_meetup you wrote.
You can hover over these two objects and find that the vscode prompt the type can be any.

save() doesn't work in Mongoengine

I'm trying to perform a simple insert operation using Mongoengine and Django.
Regarding my project structure simply I have a project, AProject and an app, AnApp. I have a running mongo in a remote machine with an IP of X.X.X.X. I am able to insert document using Robomongo in it.
I have removed the default Database configuration part of the settings.py located inside the AProject directory. The newly added lines are shown below:
# Database
# https://docs.djangoproject.com/en/1.11/ref/settings/#databases
import mongoengine
# ----------- MongoDB stuff
from mongoengine import register_connection
register_connection(alias='default', name='AProject', host='X.X.X.X')
# ----------
Now let me show the models.py and the views.py located inside AnApp.
models.py
from mongoengine import Document, StringField
class Confession(Document):
confession = StringField(required=True)
views.py
from django.http import HttpResponse
from models import Confession
from mongoengine import connect
def index(request):
connect('HacettepeItiraf', alias='default')
confession = Confession()
confession.confession = 'First confession from the API'
print(confession.confession + ' Printable') # The output is --First confession from the API Printable--
print(confession.save()) # The output is --Confession object--
return HttpResponse(request)
The urls.py located inside AProject is simply as below:
from django.conf.urls import url, include
from django.contrib import admin
urlpatterns = [
url(r'^confessions/', include('confession.urls')),
url(r'^admin/', admin.site.urls),
]
When I enter http://127.0.0.1:10000/confessions/ I see a blank screen which I expect. However there is nothing saved from the API. I get the expected output except Confession object.
How can I solve this problem?

EDIT:
I found the concrete proof that currently MongoEngine's support for Django is unstable and it is corresponding to Django version 1.9:
MongoEngine documentation on Django support states:
Django support has been split from the main MongoEngine repository. The legacy Django extension may be found bundled with the 0.9 release of MongoEngine.
and
6.1. Help Wanted!
The MongoEngine team is looking for help contributing and maintaining a new Django extension for MongoEngine! If you have Django experience and would like to help contribute to the project, please get in touch on the mailing list or by simply contributing on GitHub.
Which leads to this repo, where the following is stated:
THIS IS UNSTABLE PROJECT, IF YOU WANT TO USE IT - FIX WHAT YOU NEED
Right now we're targeting to get things working on Django 1.9
So it may be possible that it cannot play well with Django at this state or with your version and thus the problem occurs.
Initial attempt, leaving it here for legacy reasons.
I believe that the problem occurs on how you are initializing your object, although I do not have a set up to test this theory.
It is generally considered that is better to make a new object with the .create() method:
def index(request):
connect('HacettepeItiraf', alias='default')
confession = Confession.objects.create(
confession='First confession from the API'
)
print(confession.confession + ' Printable')
confession.save()
return HttpResponse(request)
Have a look at the Django & MongoDB tutorial for more details.
In this tutorial, the supported Django version is not mentioned, but I haven't found concrete proof that MongoDB Engine can or can't play well with Django version > 1.8.
Good luck :)

Password protect a whole django app

I am running a simple staging env on heroku and I am now looking to password protect the whole app with some sort of simple authentication
I am wondering if there is a simple app or middleware that already supports this.
Have tried looking around for solutions with Heroku / Cloudflare and django, but nothing seems really straight forward.
Django 1.3.1

I use django-lockdown for exactly this purpose. It allows you to add a simple password over the whole of a dev site, without having to add in any extra auth bits on your views that aren't used outside of a dev environment. It also means you can login as admin, or regular users to test whatever your site does
https://github.com/Dunedan/django-lockdown
I use Heroku and Lockdown with this bit of code in my settings.py file
USE_LOCKDOWN = os.environ.get('USE_LOCKDOWN', 'False') == 'True'
if USE_LOCKDOWN:
INSTALLED_APPS += ('lockdown',)
MIDDLEWARE_CLASSES += ('lockdown.middleware.LockdownMiddleware',)
LOCKDOWN_PASSWORDS = (os.environ.get('LOCKDOWN_PASSWORD', 'False'),)
LOCKDOWN_URL_EXCEPTIONS = (r'^/some/url/not/locked/down/$',)
Then obviously set a config var of USE_LOCKDOWN as True on my dev site, and False on my production site so no need to change the code for either.

Django's authentication framework has built-in utilities #login_required which helps you to password protect your view functions (and its corresponding "url" of course).
Usage like this:-
from django.contrib.auth.decorators import permission_required, login_required
#login_required
def view_organization(request, org_slug):
"""
Only permit a logged in user to view the organization.
"""
org = get_object_or_404(organization, slug=org_slug)
org_users = organizationuser.objects.filter(organization=org,\
organization__is_active=true)
template = 'organizations/view_organization.html'
template_vars = {'org_users': org_users, 'org': org}
return render(request, template, template_vars)
For advanced access control, use #permission_required decorator.

Recaptcha in Django without forms?

I am putting together a Django app that has forms on the site, but I am rendering those forms manually (i.e. actually typing out each field and submitting them with AJAX).
How can I integrate Recaptcha into my forms? Thanks for the help!

recaptcha-client doesn't work with python3. I ended up using django-recaptcha (https://pypi.python.org/pypi/django-recaptcha/1.0). The brief documentation explains how to implement recaptcha using the formfield 'ReCaptchaField', but you can just use the submit function from captcha.client like this:
import captcha.client
[...]
recaptcha_response = captcha.client.submit(
request.POST.get('recaptcha_challenge_field'),
request.POST.get('recaptcha_response_field'),
'[[privatekey]]',
request.META['REMOTE_ADDR'],)
Then you check whether recaptcha_response.is_valid.
No need to add recaptcha to your INSTALLED_APPS or anything.

I am just using this python client for recaptcha:
http://pypi.python.org/pypi/recaptcha-client
then my view looks like this:
captcha_key = get_config('RECAPTCHA_PUB_KEY',None)
recaptcha_challenge_field = request.POST.get('recaptcha_challenge_field', None)
recaptcha_response_field = request.POST.get('recaptcha_response_field', None)
check_captcha = captcha.submit(recaptcha_challenge_field, recaptcha_response_field, settings.RECAPTCHA_PRIVATE_KEY, request.META['REMOTE_ADDR'])
if check_captcha.is_valid is False:
log.info('captcha_error : %s' % check_captcha.error_code)
return {'TEMPLATE':template_name,'captcha_error': True,'register_form': f,'captcha_key':captcha_key ,'next':redirect_to}

Django RSS Feed Authentication

I am looking into adding RSS feeds to one of my Django apps and I would like to be able to have them be authenticated.
I want to use the new syndication framework in Django 1.2. I've read the docs on how to do this and have the basic feeds setup.
I am new to authenticating feeds, so I am not sure what the best approach to take is or what my options really are.
Each user has a unique sub domain and I would like the URL structure to look something like this: http://mysubdomain.mysite.com/myapp/rss/ if possible.
I don't want the feeds to be publicly available, is it possible to use the users username and password for the authentication? Have you found that most feed readers support this? If it's not possible to authenticate for each user, should I try to use a uuid to give them a unique url or is that not secure enough?
As you can probably tell I am not sure what direction to take with this, so any advice on the best way to do this would be very much appreciated.
Thanks

This is an old thread, but I recently encountered the same question. I solved it by overloading the __call__ method of the Feed object:
from django.http import HttpResponse
class ArticleFeed(Feed):
"snip [standard definitions of title, link, methods...]"
def __call__(self,request,*args,**kwargs):
if not request.user.is_authenticated():
return HttpResponse(status=401)
else:
return super().__call__(request,*args,**kwargs)

Have you tried wrapping the syndication view django.contrib.syndication.views.feed into a view that requires login? RSS feeds should normally be fetched over HTTP, so this should work!
# Import Django's standard feed view.
from django.contrib.auth.decorators import login_required
from django.django.contrib.syndication.views import feed
# Wrap it in a new feed view that requires authentication!
private_feed = login_required(feed)
Caveat: I've never tried this!
Edit!
To be safe with RSS readers that don't support redirection, return a HTTP 401 status code with the following:
authentication_url = '/accounts/login'
def feed_safe_login_required ( view ):
def _ ( request, *args, **kwargs ):
if not request.user.is_authenticated:
return HttpResponseNotAuthorized, authentication_url
return _
feed = feed_safe_login_required(django.contrib.syndication.views.feed)
Where HttpResponseNotAuthorized is as defined in this django snippet.