Django community: RSS
This page, updated regularly, aggregates Django Q&A from the Django community.
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Launching blender console docker container with Django
I'm making a web application where the user can render an image with blender. I'm using django as the framework and have it running in a docker environment. My question is what would be the best practice to launch blender console commands? and more specifically how do I launch an docker container with arguments that includes the blender console. -
When i execute unittest command "python3 -m unittest filepath/filename.py" it showing improperly configured error
django.core.exceptions.ImproperlyConfigured: Requested setting INSTALLED_APPS, but settings are not configured. You must either define the environment variable DJANGO_SETTINGS_MODULE or call settings.configure() before accessing settings. -
How to delete Django migration files in Dokku?
In a normal world, you can access your application folder and delete files and folders. For Django migration files, the following code can be applied: find . -path "*/migrations/*.py" -not -name "__init__.py" -delete find . -path "*/migrations/*.pyc" -delete But in Dokku, everything is in a container that I have no idea how to access. So how can I delete migration files? -
Model Design in Django
I have many components and I want to have for each component a list where it shows users have taken how many component class Component(models.Model): name=models.CharField(max_length=128,unique=False,blank=False) detail=models.TextField() max_num=models.IntegerField(default=0) issued_num=models.IntegerField(default=0) def __str__(self): return self.name def save(self,*args,**kwargs): super().save(args,kwargs) def available(self): return self.max_num-self.issued_num Status=((0,"Pending"),(1,"Accepted"),(2,"Rejected")) class Issued(models.Model): request_user=models.ForeignKey(User,on_delete=models.CASCADE) component=models.ForeignKey(Component,on_delete=models.CASCADE) status=models.IntegerField(choices=Status,default=0) request_num=models.IntegerField(default=0) def __str__(self): return self.component.name I have come up with this solution where Issued model depending on status will be shown as request or will be treated confirmation for issue request depending on its status Can anyone please tell whether its right way to do it or if its not suggest database friendly design -
Django view html to pdf executed twice
I have following Class to generate a pdf file where I use django-renderpdf to generate a pdf from a html template. But the view is executed twice and an error is thrown. My class: class WeeklyMetre(PDFView): template_name = 'reports/invoice/weekly_metre.html' allow_force_html = True prompt_download = True @property def download_name(self) -> str: invoice = Invoice.objects.get(pk=self.kwargs['pk']) return f"WeeklyMetre_{invoice.invoice_number}.pdf" def get_context_data(self, *args, **kwargs): context = super().get_context_data(*args, **kwargs) invoice = Invoice.objects.get(pk=self.kwargs.get('pk')) market_labor_specifications = _getWeeklyMetreData(invoice=invoice) # calculate reported items: reported market_labor_specifications # invoiced specifications which are validated in invoice-period # but labor_date before invoice-period reported_mls = MarketLaborSpecification.objects.filter(invoice_id=self.kwargs.get('pk'), market_labor__labor_date__lt=invoice.period_from) \ .values('market_labor__labor_date', 'specification__position', 'specification__name') \ .order_by('market_labor__labor_date', 'specification__position', 'specification__name') \ .annotate(sum_pos=Sum('validated_quantity')) context.update({ 'invoice': invoice, 'market_labor_specifications': market_labor_specifications, 'reported_mlss': reported_mls }) print('context data', datetime.datetime.now()) return context Between the two excutions I have following error: Exception happened during processing of request from ('127.0.0.1', 57654) Traceback (most recent call last): File "/usr/lib/python3.6/socketserver.py", line 654, in process_request_thread self.finish_request(request, client_address) File "/usr/lib/python3.6/socketserver.py", line 364, in finish_request self.RequestHandlerClass(request, client_address, self) File "/usr/lib/python3.6/socketserver.py", line 724, in __init__ self.handle() File "/home/t3tr4ktys/python-virtual-environments/BillOfQuantities/lib/python3.6/site-packages/django/core/servers/basehttp.py", line 174, in handle self.handle_one_request() File "/home/t3tr4ktys/python-virtual-environments/BillOfQuantities/lib/python3.6/site-packages/django/core/servers/basehttp.py", line 182, in handle_one_request self.raw_requestline = self.rfile.readline(65537) File "/usr/lib/python3.6/socket.py", line 586, in readinto return self._sock.recv_into(b) ConnectionResetError: [Errno 104] Connection reset by peer First of all I don't know why it … -
django deploy with gunicorn is worked except static but not work with nginx conf
I'd like to try deploy django server with nginx. however, it did not work. my django setting setting.py ALLOWED_HOSTS = ['my_private_IP','my_public_IP','*'] ... STATIC_URL = '/home/ubuntu/static/' while I execute gunicorn command with config, it worked except static file. gunicorn -c conf/gunicorn_config.py myproject.wsgi gunicorn_config.py conmmand = '/home/ubuntu/venv/bin/gunicorn' pythonpath = '/home/ubuntu/myproject' bind = 'my_private_IP:8069' Then I'd like to deal with static file. After setting nginx conf, the connection kept waiting. This is my nginx conf in /etc/nginx/sites-available and enable server{ listen 80; server_name _; location /static/ { root /home/ubuntu/static/; } location / { include proxy_params; proxy_pass http://my_private_IP:8069; } } I have changed server_name serveral times (localhost, 127.0.0.1, my_private_IP ...), whatever it was, the connection was still stuck. How can fix to connect to the server? And Is my static conf right? Thanks a lot. -
I am unable to load static files in django although i have done everything
in my settings.py file i did STATICFILES_LOCATION='static' STATIC_URL = '/static/' STATICFILES_DIR=[ os.path.join(BASE_DIR,'Kevin/static') ] STATIC_ROOT='static' but in my templates i am unable to load the files the template folder is in root directory whereas static folder is in django root folder named as Kevin which is also in that base folder in my html template i did the {% load static %} i am sharing some code of my template as well so you can have a look {% load static %} <meta charset="utf-8"> <meta http-equiv="X-UA-Compatible" content="IE=edge"> <meta name="viewport" content="width=device-width, initial-scale=1, shrink-to-fit=no"> <meta name="description" content=""> <meta name="author" content=""> <title>Kevin - Dashboard</title> <!-- Custom fonts for this template--> <link href="{% static 'vendor/fontawesome-free/css/all.min.css'%}" rel="stylesheet" type="text/css"> <link href="https://fonts.googleapis.com/css?family=Nunito:200,200i,300,300i,400,400i,600,600i,700,700i,800,800i,900,900i" rel="stylesheet"> <!-- Custom styles for this template--> <link href="{%static 'css/sb-admin-2.min.css'%}" rel="stylesheet"> -
node.JS / Django - configure Apache / NGINX to run from location
I am using Apache as my frontend VPS / NGINX as backend VPS for node.js / django apps with react.js frontend. I am trying to run my apps from same VPS, but from different locations: e.g.: https://myapps/app1 https://myappps/app2 Previously I have tackled this issue by adding the location in the router of the apps - however I am not sure if this is the best way to do this. Can I actually add the location in the VPS config of NGINX or Apache? # nginx.default server { listen 8020; server_name example.org; location / { proxy_pass http://127.0.0.1:5000; proxy_set_header Host $host; proxy_set_header X-Forwarded-For $proxy_add_x_forwarded_for; } location /static { root /opt/app/client/build/; } } -
getting JSON of uploaded pdf file in Django
I have uploaded a pdf file in django server from html page, now I want to show the JSON of same file on HTML page in response. Someone please help me in this. -
Querying model field data in template/views without get_object_or_404()
Homefeed is the page where i query all the blogposts In this project, any user that sees a blogpost that they are interest in can submit their interest to the post. 1 user can only submit 1 interest to that blogpost, but they can submit as many interest as they want to different blogposts. Right now in my home.html, I am trying to make it such that if YOU have submitted interest,(aka your interest status is at pending or accept or decline) for that particular blog post, you will see the view interest button instead of the submit interest button. But I am facing a problem because in my views, I am querying for blog_posts = BlogPost.objects.all() and not blog_post = get_object_or_404(BlogPost, slug=slug). As such, how am I able to query whether or not for the particular blogpost, the user has already submitted an interest in my template to determine which button should show in my home.html? Thanks, and also I dont want to change the url at all :) views.py def home_feed_view(request, *args, **kwargs): context = {} blog_posts = BlogPost.objects.all() context['blog_posts'] = blog_posts page = pageFilter(request.GET, queryset=BlogPost.objects.exclude(author_id=request.user.id).order_by('date_updated')) context['page'] = page paginated_page = Paginator(page.qs, 4) page = request.GET.get('page') page_obj = … -
how can i open a modal from views.py in django? Is there any way?
I have a django application. when a user clicks the submit button two modals should be shown one after another each has to do different functions in views.py in my django app. I am trying to write the code but the problem is not easy and I have looked everywhere and couldn't get the answer. I have to write the code in the following way: views.py def submit(request): #first modal should pop up which has two options. #secound modal should pop up which also has a two options. and when the user clicks any of the buttons each should also call the different function in views.py. html <!DOCTYPE html> <html lang="en"> <head> <meta charset="UTF-8"> <title>Title</title> </head> <body> <button class="btn btn-danger custom" style="position:absolute; left:800px; top:180px;" type="submit">Update the data</button> # user clicks a button in form and modal should pop up # then 2 modals should pop up # the first modal has two buttons # when user clicks the update button the function in the views should be called and also 2nd modal should pop up <!--modal 1--> <div aria-hidden="true" aria-labelledby="exampleModalLabel" class="modal fade" id="UpdateModal" role="dialog" tabindex="-1"> <div class="modal-dialog" role="document"> <div class="modal-content"> <div class="modal-header"> <h5 class="modal-title" id="exampleModalLabel">Update Changes</h5> <button aria-label="Close" class="close" data-dismiss="modal" … -
is it possible to create the link of selenium file and insert jquery?
I have created a Python Selenium script in Django framework. I want to manage the link of my selenium project. So that whenever I hit the link in a browser, it should run testing. I don't want to run it from the terminal, as I want to add jquery. here is my code driver = webdriver.Chrome(executable_path='/home/sr/PycharmProjects/selenium/automation/KPTGR/chromedriver') driver.get("https://www.google.com/") get = driver.find_element_by_id("gsr") print(get.text) -
How to short down this code to create django model so it works the same in term of functionality
Just imagine Netflix Site, I want to store video links of multiple seasons of TV shows with multiple episodes, I want to increment number of seasons and episode of a certain tv show on the need specifically from admin area. I don't know ho to do that instead of just creating variable of every episode link and season model. Please Help ! Thanx in adv class Tv_Shows(models.Model): STATUS_CHOICES = ( ('draft', 'Draft'), ('published', 'Published'), ) title = models.CharField(max_length=250) slug = models.SlugField(max_length=250, unique_for_date='publish') author = models.ForeignKey(User, on_delete=models.CASCADE, related_name='blog_posts') body = models.TextField() episode1 = models.CharField(max_length=100, null=True) episode2 = models.CharField(max_length=100, null=True) episode3 = models.CharField(max_length=100, null=True) episode4 = models.CharField(max_length=100, null=True) episode5 = models.CharField(max_length=100, null=True) episode6 = models.CharField(max_length=100, null=True) episode7 = models.CharField(max_length=100, null=True) episode8 = models.CharField(max_length=100, null=True) episode9 = models.CharField(max_length=100, null=True) episode10 = models.CharField(max_length=100, null=True) imdb_rating = models.DecimalField(null=True, max_digits=12, decimal_places=1) publish = models.DateTimeField(default=timezone.now) created = models.DateTimeField(auto_now_add=True) updated = models.DateTimeField(auto_now=True) views = models.IntegerField(default=0) status = models.CharField(max_length=10, choices=STATUS_CHOICES, default='draft') class Meta: ordering = ('-publish',) def __str__(self): return self.title def get_absolute_url(self): return reverse('blog:post_detail', args=[self.publish.year, self.publish.month, self.publish.day, self.slug]) objects = models.Manager() # The default manager. published = PublishedManager() # Our custom manager. tags = TaggableManager() -
How to set up a homepage at django
I am new to django and I want to set my homepage url My homepage is 127.0.0.1:8000/home but I want my homepage at 127.0.0.1:8000/ or I want my django start as 127.0.0.1:8000/home as default. How do I do it? -
unable to upload multi files through django serializer
I can upload single files at a time but cant upload multiple files at once. Any idea how can I upload multifiles ? my models.py class VideoPost(models.Model): user = models.ForeignKey("profiles.HNUsers", on_delete=models.DO_NOTHING) video = models.FileField("Post Video", blank=True, null=True) timestamp = models.DateTimeField("Timestamp", blank=True, null=True, auto_now_add=True) text = models.TextField("Description text", blank=True) class Meta: verbose_name_plural = "VIdeo Posts" my serializers.py class VideoPostSerializer(serializers.ModelSerializer): class Meta: model = VideoPost fields = ( 'user', 'text', 'video', ) and my views.py def video_post(request): if request.method == 'POST': data = request.data print(data) serializer = VideoPostSerializer(data=data , many =True) if serializer.is_valid(): serializer.save() print("audio object saved") try: video_post_object = VideoPost.objects.filter(user__id=data['user']).order_by('-timestamp')[0] print(video_post_object) try: post = Posts() #testing multi files post.user = HNUsers.objects.get(id=data['user']) post.audio_url = video_post_object.video.url post.type = 'V' post.category = data['category'] post.created_on = video_post_object.timestamp post.text = video_post_object.text save_post = post.save() post_id = post.pk try: user = HNUsers.objects.get(pk=data['user']) if user.user_type == 'HN': payload = { "user_name": user.full_name, "image_url": user.profile_img_url, "post_id": post_id, "notification_type": "hn" } print(payload) broadcast_notification('all', payload, 1) except HNUsers.DoesNotExist: print("user_id_does_not_exist") print("post object created") except Posts.DoesNotExist: print("Failed - post creation failed") except HNUsers.DoesNotExist: print("Failed - user object not found") except AudioPost.DoesNotExist: print("Failed - video object not found") return Response(serializer.data, status=status.HTTP_200_OK) -
Celery: Sending email using the Celery throws "Invalid address" error
I am using Django + Celery to send emails. I am basically new to Celery. Whenever I try to send emails using the tasks.py file it shows me following error on console. backends/smtp.py", line 122, in _send recipients = [sanitize_address(addr, encoding) for addr in email_message.recipients()] **File "/home/ropali/Development/PythonWorkSpace/hrms_venv/lib/python3.9/site-packages/django/core/mail/**backends/smtp.py", line 122, in recipients = [sanitize_address(addr, encoding) for addr in email_message.recipients()] File "/home/ropali/Development/PythonWorkSpace/hrms_venv/lib/python3.9/site-packages/django/core/mail/message.py", line 84, in sanitize_address raise ValueError('Invalid address "%s"' % addr)ValueError: Invalid address "['test2@test.com']" My tasks.py file import string from django.contrib.auth.models import User from django.utils.crypto import get_random_string from celery import shared_task from hrms.services import EmailService from django.conf import settings @shared_task def send_marketing_email(email, subject): email_service = EmailService(subject=subject, to=email) # email_service.set_to(email) email_service.load_template('templates/marketing_email.html',context={ 'url': settings.BASE_URL + 'pages/contact-us' }) email_service.send() return 'Email sent to {}'.format(email) My view code: class SendEmailsView(LoginRequiredMixin, AdminRequired,View): template_name = 'marketing/send_emails.html' context = {} def get(self, request): return render(request, self.template_name, self.context) def post(self, request): subject = request.POST.get('subject') if not subject: return JsonResponseBuilder.error('Subject field is required.',HTTPStatus.UNPROCESSABLE_ENTITY.value) # get all the names & emails from the uploaded excel file # send the email to each email address from the file email_df = pd.read_csv(settings.MEDIA_ROOT / constants.EMAIL_LEADS_EXCEL_FILE) for index,row in email_df.iterrows(): send_marketing_email.delay(row['email'], subject) return JsonResponseBuilder.success('Emails sent successfully...') services.py file which contains EmailService class. class EmailService(): email … -
Add Django Rest Framework route to third party urls
I am using Djoser along with Django REST Framework. This is my urls.py: from django.urls import path, include from rest_framework.routers import DefaultRouter djoser_urls = [ path("auth/", include("djoser.urls")), path("auth/", include("djoser.urls.authtoken")), ] urlpatterns = router.urls + djoser_urls When using the browsable API at localhost:8000/auth/ the djoser urls are available. So that is all working correctly. What I need is to be able to browse to localhost:8000/ and have the above route displayed as an option to click on. Currently there are no routes displayed in the browsable API at localhost:8000/ even though if I manually type localhost:8000/auth/ into the url then I am taken to the djoser urls. In other words, when opening the browsable API at localhost:8000/, I would like the following to be available to click on: { "auth": "http://localhost:8000/auth/" } -
What is the reason for Django- Page not found /login error
I am working on a Django project,upon running the application with python3 manage.py runserver. I am able to get the login page,after logging in,I should be able to see the home/index page however I get an error something like Request Method: POST Request URL: http://127.0.0.1:8000/login Using the URLconf defined in leadgen.urls, Django tried these URL patterns, in this order: index [name='index_view'] [name='login_view'] add [name='add_view'] base [name='base_view'] company [name='company_view'] insigths [name='insights_view'] master [name='master_view'] leads [name='leads_view'] settings [name='settings_view'] team [name='team_view'] admin/ The current path, login, didn't match any of these.``` Leadgen is my project and Leadgen_app is its application. This is how I have written the urls.py for the application leadgen_app/urls.py from django.urls import path from . import views urlpatterns = [ path('', views.login_view, name='login_view'), path('index', views.index_view, name='index_view'), path('add', views.add_view, name='add_view'), path('base', views.base_view, name='base_view'), path('company', views.company_view, name='company_view'), path('insigths', views.enrich_view, name='insights_view'), path('master', views.master_view, name='master_view'), path('leads', views.leads_view, name='leads_view'), path('settings', views.settings_view, name='settings_view'), path('team', views.team_view, name='team_view'), ] and this is how I have written the urls for leadgen project from django.contrib import admin from django.urls import path,include urlpatterns = [ path('', include('leadgen_app.urls')), path('admin/', admin.site.urls), ]``` How do I solve this error? Please help. -
Call a view from from in django
I want to call a view from my html template using forms but when i click on update button nothing happens. I cant seem to find out the error. Codes are as follows HTML TEMPLATE <form action = {% url 'update' %} method = "POST"> {% csrf_token %} <h6 class="heading-small text-muted mb-4">User information</h6> <div class="pl-lg-4"> <div class="row"> <div class="col-lg-6"> <div class="form-group"> <label class="form-control-label" for="input-username">Company ID</label> <input type="text" name="id" disabled="disabled" id="input-username" class="form-control form-control-alternative" value="{{ data.id }}"> </div> </div> <div class="col-lg-6"> <div class="form-group"> <label class="form-control-label" for="input-email">Email address</label> <input type="email" name="email" id="input-email" class="form-control form-control-alternative" value="{{ data.email }}"> </div> </div> </div> <div class="row"> <div class="col-lg-6"> <div class="form-group"> <label class="form-control-label" for="input-first-name">Company Name</label> <input type="text" name="name" id="input-first-name" class="form-control form-control-alternative" value="{{ data.name }}"> </div> </div> </div> </div> <hr class="my-4" /> <!-- Address --> <h6 class="heading-small text-muted mb-4">Contact information</h6> <div class="pl-lg-4"> <div class="row"> <div class="col-md-12"> <div class="form-group"> <label class="form-control-label" for="input-address">Address</label> <input id="input-address" name="add" class="form-control form-control-alternative" placeholder="Home Address" value="{{ hno }}" type="text"> </div> </div> </div> <div class="row"> <div class="col-lg-4"> <div class="form-group"> <label class="form-control-label" for="input-city">City</label> <input type="text" name="city" id="input-city" class="form-control form-control-alternative" placeholder="City" value="{{ city }}"> </div> </div> <div class="col-lg-4"> <div class="form-group"> <label class="form-control-label" for="input-country">Country</label> <input type="text" name="country" id="input-country" class="form-control form-control-alternative" placeholder="Country" value="{{ country }}"> </div> </div> <div class="col-lg-4"> <div class="form-group"> … -
ModelBase object argument after ** must be a mapping, not list
I have been working on certain endpoint on Rest Framework I came to following problem. ModelBase object argument after ** must be a mapping, not list In my usecase.py file I have following line of code. class AddRouteRepresentativeWithBusCompanyRouteUseCase(BaseUseCase): def __init__(self, bus_company_route: BusCompanyRoute, serializer ): self._bus_company_route = bus_company_route self._serializer = serializer self._data = self._serializer.validated_data.get('data') def execute(self): self._factory() def _factory(self): print(self._data) route_representative = RouteRepresentative( **self._data, bus_company_route=self._bus_company_route, ) try: route_representative.full_clean() except DjangoValidationError as e: raise ValidationError(e.message_dict) RouteRepresentative.objects.bulk_create(route_representative) Seems I am getting error on **self._data. as I got orderedDict from serializer fields. What should I do in **self.data instead? TIA -
How to get subcategory with their category in django models and showing in templates
I want to present my skill category with their subcategories list. I am facing issue while fetching the data from models to views. my code is given below: model.py Here I create 2 classes SkillCategory and SkillSubCategory class SkillCategory(models.Model): id = models.AutoField(primary_key=True) category = models.CharField(max_length=100) def __str__(self): return self.category class SkillSubCategory(models.Model): id = models.AutoField(primary_key=True) category = models.ForeignKey(SkillCategory, on_delete=models.CASCADE) subcategory = models.CharField(max_length=100) icon_name = models.CharField(max_length=100) def __str__(self): return self.subcategory views.py Here I define a function based view named as skill which return list of list def skill(request): allsubcat = [] skillcat = SkillSubCategory.objects.values() cat = {c['subcategory'] for c in skillcat} for c in cat: skill = SkillSubCategory.objects.filter(subcategory=c) allsubcat.append([skill]) ctx = {'allsubcat':allsubcat} return render(request, "my_folio/skill.html", ctx) skill.html This is the template which should present the Skill like: Skill Category 1 Skill SubCat 1 Skill SubCat 2 Skill Category 2 Skill SubCat 1 Skill SubCat 2 Skill SubCat 3 Could anyone please help me for the given issue. Thanks and cheers!! -
Django Admin does not work after deployment Heroku
After deploying my application the admin of the website does not work. I am able to use the website, create an account and do all the crud, but the admin does not work. 1 - I created the superuser with python manage.py createsuperuser 2 - I deployed and tried to access the admin https://djangotodowoo.herokuapp.com/admin Although, every time that I try to access this page it redirects to: https://djangotodowoo.herokuapp.com/admin/login/?next=/admin/ I am not sure if that's the problem, but it's a hint at least. This is happening because I have on my settings.py the following line LOGIN_URL = '/login' That checks if someone is trying to access a page without login and redirects to the login page. I am not sure what the problem actually is, any ideas on how to solve the issue? Thank you very much folks! -
Run R script in Django
I'm building a site which contains lot of optimization work and all those work are done in R language. I'm building whole site on Django. Is there any way I can connect both R code and Django or what should I do here ? I tried installing rpy2 which works fine in python but in Django I'm getting a error: File "F:\django-app\restapi\views.py", line 25, in <module> import rpy2.robjects as robjects File "F:\django-app\venv\lib\site-packages\rpy2\robjects\__init__.py", line 19, in <module> from rpy2.robjects.robject import RObjectMixin, RObject File "F:\django-app\venv\lib\site-packages\rpy2\robjects\robject.py", line 6, in <module> rpy2.rinterface.initr() File "F:\django-app\venv\lib\site-packages\rpy2\rinterface\__init__.py", line 208, in initr initr(r_preservehash=r_preservehash) RuntimeError: R_USER not defined. what's the best option to do this. Is there any thing I'm missing here? -
Why can't I remove the member from the list and delete their interest?
Every blogpost has their own members. To become a member, the interest you submit must be accepted. i am trying to build a remove member function whereby if you are a member, you can remove yourself from the list and if you are the blog_post.author, you can remove ANYONE except yourself (because to remove yourself you need to delete the blogpost entirely. Once you are removed from the member-list, I also want the initial interest they have submitted to be deleted. I have attached my models.py, urls.py, template.html and views.py in this question. First, i am getting no errors, but I'm unsure of why the interest is not deleted and the member is also not removed. Is there anything wrong with my code? models.py class Account(AbstractBaseUser): email = models.EmailField(verbose_name="email", max_length=60, unique=True) username = models.CharField(max_length=30, unique=True) class BlogPost(models.Model): title = models.CharField(max_length=50, null=False, blank=False, unique=True) author = models.ForeignKey(settings.AUTH_USER_MODEL, on_delete=models.CASCADE) slug = models.SlugField(blank=True, unique=True) members = models.ManyToManyField(settings.AUTH_USER_MODEL, blank=True, related_name="members") class Interest(models.Model): user = models.ForeignKey(settings.AUTH_USER_MODEL, on_delete=models.CASCADE) blog_post = models.ForeignKey(BlogPost, on_delete=models.CASCADE) class InterestInvite(models.Model): ACCEPT = "ACCEPT" DECLINE = "DECLINE" PENDING = "PENDING" STATUS_CHOICES = [ (ACCEPT, "accept"), (DECLINE, "decline"), (PENDING, "pending"), ] interest = models.OneToOneField(Interest, on_delete=models.CASCADE, related_name="interest_invite") status = models.CharField(max_length=25, choices=STATUS_CHOICES, default=PENDING) views.py class InterestMixin(object): … -
Can´t excecute view action in Django
I would like to understand why when I click on login button I can't run user_login function. I'm new in django. Thanks urls.py: from django.urls import path from iol_dash import views urlpatterns = [ path('', views.index, name='index'), ] views.py: from django.shortcuts import render from iol import Service def index(request): return render(request, 'index.html', {}) def user_login(request): username = request.POST.get('username') password = request.POST.get('password') service = Service() token = service.get_token( username=username, password=password ) print(token) index.html: <form class="modal-content animate" method="post" action={{ views.user_login }}> {% csrf_token %} {{ form.as_p }} <div class="imgcontainer"> <span onclick="document.getElementById('id01').style.display='none'" class="close" title="Close Modal">&times;</span> </div> <div class="container"> <label for="username"><b>Username</b></label> <input type="text" placeholder="Enter Username" name="username" required> <label for="password"><b>Password</b></label> <input type="password" placeholder="Enter Password" name="password" required> <button type="submit">Login</button> </div> </form>
