Django community: RSS
This page, updated regularly, aggregates Django Q&A from the Django community.
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Logging in Python Django: INFO not logging onto file
I have setup a logger in my Django project for 4 different cases: info messages, debug messages, error messages, ad gunicorn logging. Here is the content of my settings.py: LOGGING = { "version": 1, "disable_existing_loggers": False, "formatters": { "main_formatter": { "format": "{asctime}-{levelname}-{module}-{funcName}-{message}", "style": "{", }, }, "handlers": { "console": { "class": "logging.StreamHandler", "formatter": "main_formatter", }, "dfile": { "class": "logging.FileHandler", "filename": "logs/debug.log", "formatter": "main_formatter", }, "file": { "class": "logging.FileHandler", "filename": "logs/info.log", "formatter": "main_formatter", }, "efile": { "class": "logging.FileHandler", "filename": "logs/error.log", "formatter": "main_formatter", }, "gfile": { "class": "logging.FileHandler", "filename": "logs/gunicorn.log", "formatter": "main_formatter", }, }, "loggers": { "main": { "handlers": ["dfile", "console"], "propagate": True, "level": "DEBUG", }, "main": { "handlers": ["file", "console"], "propagate": True, "level": "INFO", }, "main": { "handlers": ["efile", "console"], "propagate": True, "level": "ERROR", }, "gunicorn.access": { "handlers": ["gfile", "console"], "propagate": False, "level": "DEBUG", }, }, } And here is an example: import logging [...] logger = logging.getLogger(__name__) [...] logger.error(f"{request.user}: Erreur interne. Code [LI.001]") # THIS WORKS logger.info(f"Password Generated and Sent to {email}") # THIS DOESNT WORK In my logs/ folder, I have 4 files: info.log, error.log, debug.log and gunicorn.log The only output I see is in error.log and gunicorn.log, the other 2 files are always empty. -
Cannot import settings module in django
I am building a user interface with the help of Django's authentication infrastructure. However, when i try to import any module (such as the 'User' model or the 'authenticate' function) that needs access to the settings module, i get such error: c:/Users/Lorenzo/Desktop/progetto_mondodb_di_lorenzo_giaretta/Bit4All/BitEx/main.py Traceback (most recent call last): File "c:\Users\Lorenzo\Desktop\progetto_mondodb_di_lorenzo_giaretta\Bit4All\BitEx\main.py", line 4, in <module> from views import mainWindow File "c:\Users\Lorenzo\Desktop\progetto_mondodb_di_lorenzo_giaretta\Bit4All\BitEx\views.py", line 3, in <module> from models import Profile File "c:\Users\Lorenzo\Desktop\progetto_mondodb_di_lorenzo_giaretta\Bit4All\BitEx\models.py", line 2, in <module> from django.contrib.auth.models import User File "C:\Users\Lorenzo\Desktop\progetto_mondodb_di_lorenzo_giaretta\myvenv\lib\site-packages\django\contrib\auth\models.py", line 2, in <module> from django.contrib.auth.base_user import AbstractBaseUser, BaseUserManager File "C:\Users\Lorenzo\Desktop\progetto_mondodb_di_lorenzo_giaretta\myvenv\lib\site-packages\django\contrib\auth\base_user.py", line 47, in <module> class AbstractBaseUser(models.Model): File "C:\Users\Lorenzo\Desktop\progetto_mondodb_di_lorenzo_giaretta\myvenv\lib\site-packages\django\db\models\base.py", line 107, in __new__ app_config = apps.get_containing_app_config(module) File "C:\Users\Lorenzo\Desktop\progetto_mondodb_di_lorenzo_giaretta\myvenv\lib\site-packages\django\apps\registry.py", line 252, in get_containing_app_config self.check_apps_ready() File "C:\Users\Lorenzo\Desktop\progetto_mondodb_di_lorenzo_giaretta\myvenv\lib\site-packages\django\apps\registry.py", line 134, in check_apps_ready settings.INSTALLED_APPS File "C:\Users\Lorenzo\Desktop\progetto_mondodb_di_lorenzo_giaretta\myvenv\lib\site-packages\django\conf\__init__.py", line 79, in __getattr__ self._setup(name) File "C:\Users\Lorenzo\Desktop\progetto_mondodb_di_lorenzo_giaretta\myvenv\lib\site-packages\django\conf\__init__.py", line 66, in _setup self._wrapped = Settings(settings_module) File "C:\Users\Lorenzo\Desktop\progetto_mondodb_di_lorenzo_giaretta\myvenv\lib\site-packages\django\conf\__init__.py", line 144, in __init__ mod = importlib.import_module(self.SETTINGS_MODULE) File **"C:\Users\Lorenzo\AppData\Local\Programs\Python\Python39\lib\importlib\__init__.py", line 127, in import_module return _bootstrap._gcd_import(name[level:], package, level) ModuleNotFoundError: No module named 'Bit4All'** It looks like in the _init_.py file, inside django/conf, which currently resides inside my virtual environment, something wrong happens while trying to import the setting module (with importlib.import_module). The snippet of code responsible for the problem: class Settings: def __init__(self, settings_module): … -
How to view all Feature Flags in views.py
I have added feature flags in my django. I defined a switch in django admin panel. The default Switch model provides all the necessary functionality.To list all the switches using manage.py, we use this command: ./manage.py waffle_switch -l How to list all the switches in views.py? -
Automatic documentation of api roles and permissions (django rest framework)
I have a Django project with various REST APIs. The permission to access those APIs are defined throught rest_framework permission classes and django.contrib.auth.models Permission model. The documentation of those apis is automatically generated using drf_yasg library, i need to find a way to inlcude in the schema generated from drf-yasg the needed permission to access every api without the need to write id down manually. Anyone can give me an hint? -
Django save form with foreign key
I am currently trying to create a form where users get to fill in their details after creating an account. The idea is that every user, after registration, gets redirected to this form page to fill it out. To achieve this, i'm using foreign keys.However it doesn't save to database. models.py class User(AbstractUser): pass def __str__(self): return self.username class Detail(models.Model): user = models.ForeignKey(User, on_delete=models.CASCADE, null=False, default="") first_name = models.CharField(max_length=200, default="") last_name = models.CharField(max_length=255, default="") class Meta: verbose_name_plural = "Detail" def __str__(self): return self.first_name+ " "+self.last_name forms.py class Details(forms.ModelForm): class Meta: model = Detail fields = "__all__" widgets={ "user": forms.TextInput() } views.py def details(request): if request.method =="POST": form = Details(request.POST) if form.is_valid(): detail = form.save(commit=False) detail.user = request.user detail.first_name = detail.first_name.lower() detail.last_name = detail.last_name.lower() detail.save() return redirect("admin:index") else: form = Details(initial={"user":request.user.username}) return render(request, "details.html", {"form":form}) -
Set default author to logged in user
I have already tried this, this and this to no avail. When submitting a comment I want to default the author field to be the currently logged in user. I am using class based views and haven't found many questions and answers besides trying to implement the above. views.py class CommentPost(SingleObjectMixin, FormView): model = Article form_class = CommentForm template_name = "article_detail.html" def post(self, request, *args, **kwargs): self.object = self.get_object() return super().post(request, *args, **kwargs) def form_valid(self, form): comment = form.save(commit=False) form.instance.author = self.request.user comment.article = self.object comment.save() return super().form_valid(form) def get_success_url(self): article = self.get_object() return reverse("article_detail", kwargs={"pk": article.pk}) models.py class Comment(models.Model): article = models.ForeignKey(Article, on_delete=models.CASCADE) comment = models.CharField(max_length=140) author = models.ForeignKey( settings.AUTH_USER_MODEL, on_delete=models.CASCADE, ) def __str__(self): return self.comment def get_absolute_url(self): return reverse("article_list") article_detail.html <!-- templates/article_detail.html --> {% extends 'base.html' %} {% load crispy_forms_tags %} {% block content %} <body class="d-flex h-100"> <div class="container"> <div class="col-8 mx-auto"> <h3 class="pb-4 mb-4 fst-italic border-bottom">Article detail</h3> <div class="article-entry"> <h2>{{object.title}}</h2> <p class="text-black-50 fst-italic">by {{object.author}} | {{object.date}}</p> <p>{{object.body}}</p> </div> {% if user.is_authenticated and user == article.author or user.is_superuser %} <p> <button class="btn"> <a class="text-decoration-none text-white" href="{% url 'article_edit' article.pk %}">edit</a> </button> <button class="btn"> <a class="text-decoration-none text-white" href="{% url 'article_delete' article.pk %}">delete</a> </button> </p> {% endif %} <br/> … -
Use a form in Django running locally to get folder/file path
Task: I am running a python Django web app locally. I would like to use an input form to get the path of a folder or file for use in further programming. Problem: I don't want to have users need to copy/paste the full path string; which is inconvenient, slow and clunky. The files are large and so I do not want to upload and store a copy. Is this possible given that the app is run on the same machine as the files? -
Django DRF Api Key Override Length expiry
I can see the following on the DRF page stating the key length can be overridden but being new at Django I really dont know how to use this? https://florimondmanca.github.io/djangorestframework-api-key/guide/#creating-and-managing-api-keys This package ships with a key generation algorithm based on Django's password hashing infrastructure (see also Security). The .key_generator attribute on BaseAPIKeyManager allows you to customize key generation. For example, you can customize the length of the prefix and secret key using: from rest_framework_api_key.models import BaseAPIKeyManager from rest_framework_api_key.crypto import KeyGenerator class OrganizationAPIKeyManager(BaseAPIKeyManager): key_generator = KeyGenerator(prefix_length=8, secret_key_length=32) # Default values class OrganizationAPIKey(AbstractAPIKey): objects = OrganizationAPIKeyManager() # ... I dont understand where this would go, how it would link to the admin view, or what comes after # ... Any help would be most appreciated -
How to send Files to Another Site from a Django App?
I recently developed a django site where it is possible to upload some files. When these files are uploaded, I would like them to be send to another website. Is it possible to do this? Does it have anything to do with the FTP protocol? Thanks for your help! -
Django - Filter on ManyToManyField
I have two models: class Student(models.Model): first_name = models.CharField() last_name = models.CharField() class Group(models.Model): name = models.CharField() students = models.ManyToManyField(Student) Some data (first_name and last_name concatenated): Group #1 | Blak Coleman Group #1 | Miguel Scott Group #2 | Jordan Barnes Group #2 | Jordan Gustman Group #2 | Jekson Barnes Group #3 | Jordan Smith As you can see theres three students by name Jordan. So I need to return groups which in students queryset has only students by name Jordan. I tried this: groups = Group.objects.filter(students__first_name='Jordan') But group.first().students.all() contains all the students not only Jordan. Expected result: Group #2 | Jordan Barnes Group #2 | Jordan Gustman Group #3 | Jordan Smith How could I do this? -
ManyToMany field to same model without allowing the same object
I am using Django with a model that has a ManyToMany field to the same model. class Job(models.Model): name = models.CharField(_('Name'),max_length=80, blank=True, default="") related_jobs = models.ManyToManyField('self') this works fine. I can create Job objects and add jobs to related_jobs. The problem is that I can associate the same object to himself like this: job1 = Job.objects.create(name='Java') job2 = Job.objects.create(name='Python') job1.related_jobs.add(job2) job1.related_jobs.add(job1) #I don't want this to be possible Is there anyway to restrict this on the models? -
django-rest-framework days-of-week representation in model
I want to create a schedule model to schedule media on days of the week class Schedule(models.Model): name = models.CharField(max_length=100) start_date = models.DateTimeField() end_date = models.DateTimeField() week_days = ??? date_created = models.DateTimeField(auto_now_add=True) date_updated = models.DateTimeField(auto_now=True) status = models.BooleanField(default=True) def __str__(self): return self.name but the problem is with week_days it can have multiple days how do I design the week_days field in the Django rest framework? -
How to handle namespace API versioning with DRF-YASG
In my project we want to have multiple versions of same api so as to support backward compatibility. Right now we are using drf-yasg's swagger url at /api/doc/ While our apis are /api/vehicle/$ /api/warehouse/ And then each of the app vehicle and ware house have their own endpoints.. Now we want to do versioning as /api/v1/doc /api/v2/doc /api/v1/vehicle /api/v2/vehicle /api/v1/warehouse Goal is that /v1/doc should show v1 of both vehicle and ware house while/v2/doc should show only v2 of vehicle since only vehicle app has some apis with v2.... How to achieve this I tried adding default version v1 initially in drf settings. But that resulted in no listing being shown in swagger view -
Order queryset by the number of foreign key instances in a Django field
I am trying to return the objects relating to a through table which counts the number of reactions on a blog post. I have an Article model, Sentiment model and Reactions model. The sentiment is simply a 1 or 2, 1 representing like and 2 for dislike. On the frontend users can react to an article and their reactions are stored in a Reactions table. Reactions model class Reaction(models.Model): user_id = models.ForeignKey(User, related_name='user_id', on_delete=models.CASCADE) article_id = models.ForeignKey(Article, related_name='article_id', on_delete=models.CASCADE) sentiment = models.ForeignKey(Sentiment, related_name='sentiment', on_delete=models.CASCADE) I'd like to find the 2 most liked articles so I have written a view to handle the GET request views.py class MostPopularView(generics.RetrieveAPIView): queryset = Reaction.objects.annotate(num_likes = Count('sentiment_id')).order_by('num_likes') serializer_class = MostPopularSerializer and a serializer to transform the data serializers.py class MostPopularSerializer(serializers.Serializer): class Meta: fields = ( 'id', 'title', ) model = Article As the code stands now, I'm getting a response <QuerySet [<Reaction: d745e09b-5685-4592-ab43-766f47c73bef San Francisco Bay 1>, <Reaction: d745e09b-5685-4592-ab43-766f47c73bef The Golden Gate Bridge 1>, <Reaction: dd512e6d-5015-4a70-ac42-3afcb1747050 San Francisco Bay 1>, <Reaction: dd512e6d-5015-4a70-ac42-3afcb1747050 The Golden Gate Bridge 2>]> Showing San Francisco Bay has 2 likes and The Golden Gate Bridge has 1 like and 1 dislike. I've tried multiple methods to get the correct response including … -
Safest way to send base64 encoded Image in a post api request
I have a post request where I am sending a base64 encoded image. I wanted to know is there any encryption method that I can use to encrypt this base64 string. In frontend I am using Javascript and for backend Django is being used. So far I have not been able to find any resource or anything that can point me right direction. Any help will be appreciated. -
Django form is not saving when I have action = to desired url
If I use action="" in my django form, the form works properly but sends the user to the wrong page. I want the user to go back to the macro/ page upon form submission, but when I add that url to action (like action="{% url 'macro' %}", it goes to the page but the form doesn't save. Any suggestion on how to handle this? Code below: (Option 1) macro_update.html -> the form here works properly, but takes the user to the wrong page <ul> <form action="" method="post"> {% csrf_token %} {{ macro_form.as_ul }} <input type="submit" value="Submit"> </form> </ul> (Option 2) macro_update.html -> the user is redirected to the right page upon submission, but the form data doesn't update/save <ul> <form action="{% url 'macro' %}" method="post"> {% csrf_token %} {{ macro_form.as_ul }} <input type="submit" value="Submit"> </form> </ul> views.py @login_required(login_url='login') def macroUpdate(request): if request.method == "POST": macro_form = MacroForm(request.POST, instance=request.user.profile) if macro_form.is_valid(): macro_form.save() messages.success(request,('Your macros were successfully updated!')) else: messages.error(request,('Unable to complete request')) return redirect("profile") macro_form = MacroForm(instance=request.user.profile) context = {"user":request.user, "macro_form":macro_form } return render(request, 'macro_update.html', context) urls.py urlpatterns = [ path('', views.loginPage, name='login'), path('register/', views.registerPage, name='register'), path('profile/', views.profilePage, name='profile'), path('profile/update/', views.profileUpdate, name='profile-update'), path('logout/', views.logoutUser, name='logout-user'), path('macro/', views.macroPage, name='macro'), path('macro/update/', views.macroUpdate, name='macro-update'), … -
populating form with data from session; django
I'm wondering how to fill my form with data that i have stored in my session. my model: models.py class Order(models.Model): order_by = ForeignKey(User, on_delete=DO_NOTHING) order_status = ForeignKey(OrderStatus, on_delete=DO_NOTHING) created = DateTimeField(default=datetime.now) address_street = CharField(max_length=256) address_postal_code = CharField(max_length=18) address_city = CharField(max_length=128) shipping = ForeignKey(ShippingMethod, on_delete=DO_NOTHING) payment = DecimalField(max_digits=12, decimal_places=2, null=False) payment_method = ForeignKey(PaymentMethod, on_delete=DO_NOTHING) def __str__(self): return self.id I have a view that stores my choices in 'cart' class AddProductToCartView(View): def get(self, request, pk): cart = request.session.get("cart") if cart: for item in cart: if item["id"] == pk: item["quantity"] += 1 break else: cart.append({"id": pk, "quantity": 1}) request.session["cart"] = cart else: request.session["cart"] = [{"id": pk, "quantity": 1}] return redirect("store:cart_view") class CartView(View): def get(self, request): in_cart = [] overall_price = 0 overall_tax = 0 if "cart" in request.session: overall_price = 0 for item in request.session["cart"]: product = Product.objects.select_related("category").get(id=item["id"]) total_price = product.price * item["quantity"] tax = (total_price * product.tax) in_cart.append({ "product": product, "quantity": item["quantity"], "total_price": total_price, "tax": tax }) overall_price += total_price overall_tax += tax return render( request, template_name="cart/cart_view.html", context={ "products": in_cart if len(in_cart) > 0 else None, "overall_price": overall_price, "overall_tax": overall_tax } ) and I want to populate my form with data (some of it not everything is in my 'cart' so … -
How do I solve this input problem? (DJANGO)
I'm trying to do an autocomplete with django. In the HTML, I'm using a <select> tag to get the IDs and connect to the API for the data, anyway, that's not the focus. The truth is that this part of the select doesn't work inside the modal it should, just outside. Could anyone help me with suggestions? I tried to put it outside the modal and it worked, but I need it to work inside the modal. The code below shows where it should work. <div class="row"> <div class="col"> <strong style="font-weight: 500;" class="d-block mb-1">Pessoas a serem adicionadas</strong> <select class='form-control' id="id_pessoas" name="pessoas" required="required" data-ajax--cache="true" data-ajax--delay="250" data-ajax--type="GET" data-ajax--url='{% url "api-list-pessoa" %}' data-theme="bootstrap4" data-allow-clear="true" data-maximum-selection-length="7" data-minimum-input-length="3" data-language="pt-BR" data-placeholder="Selecione uma pessoa"> </select> </div> -
How to import custom modules in Django
I have a module created with the name metrics.py . I want to use this module in my views.py file.The command i used to import is "import metrics" in views.py file. both views.py and metrics.py are in the same folder. i am getting this error "ModuleNotFoundError: No module named 'metrics'" How to use the custom module i created in views.py -
Django JSONField is not able to encode smileys properly
I plan to store a dict in a Django JSONField. One key of this dict is a comment a user can enter. And users like a lot to add some smileys in their comments... The problem is that some smileys, are saved properly in DB. The database is MySQL 8.0.31, Django version is 4.0.8 : JSONField is supported for this environment as reported in documentation. The database default encoding is utf8mb4 and collation utf8mb4_general_ci. With that model : class TestJSONField(models.Model): data = models.JSONField() Here is the test case : comment=b'smiley : \xf0\x9f\x98\x8a'.decode() t=TestJSONField(pk=1, data={'comment':comment}) t.save() r=TestJSONField.objects.get(pk=1) print('BEFORE :', comment) print('AFTER :', r.data['comment'], '(str)') print('AFTER :', r.data['comment'].encode(), '(utf-8 encoded bytes)') which gives : BEFORE : smiley : 😊 AFTER : smiley : ? (str) AFTER : b'smiley : ?' (utf-8 encoded bytes) As you can see the smiley is not stored correctly. This smiley is 4 bytes encoded, this may be the source of the problem because with 2-bytes encoded chars I do not have any problem. With a TextField and using json dumps()/loads() I do not have any problem. Do you have an idea how to have 4 bytes encoded smileys to be saved in a JSONField ? -
logging with cookiecutter-django inside docker not displaying on console
I can't get logging to display. Am I doing something wrong? Other logging that is generated by cookiecutter-django works fine, but when I try to implement logging myself for debugging I am not able to get any display settings.py: LOGGING = { "version": 1, "disable_existing_loggers": False, "formatters": { "verbose": { "format": "%(levelname)s %(asctime)s %(module)s " "%(process)d %(thread)d %(message)s" } }, "handlers": { "console": { "level": "INFO", "class": "logging.StreamHandler", "formatter": "verbose", }, "file": { "level": "INFO", "class": "logging.FileHandler", "formatter": "verbose", "filename": "debug.log", }, }, "root": {"level": "INFO", "handlers": ["console", "file"]}, } view.py: class run(viewsets.ModelViewSet): serializer_class = RunSerializer def list(self, request): return Response("Get") def create(self, request): logger = logging.getLogger("root") logger.info(request.data) logger.info(request.POST) return Response({200}) this logging does not display to console -
How do I change the text/value of "Add [Model-Name]" button in Django Admin?
When we login to Django Admin Interface as a superuser, we see the list of models on the left sidebar. When we click on any model name, we go to the list display page of that model which have 'Add [Model-Name]" button on uuper right corner. How do I change the text/value of that button? In my case, I have User Model, and I want to change the "Add User" text on list display page of User Model to "Invite User". How do I accomplish that? I have encircled the button with red in the screenshot attached. Django Admin Interface Screenshot I have tried different solutions told in this stackoverflow question and in this django documentation. But I am unable to achieve. I tried to override change_form.html by changing {% blocktranslate with name=opts.verbose_name %}Add {{ name }}{% endblocktranslate %} to {% blocktranslate with name=opts.verbose_name %}Invite {{ name }}{% endblocktranslate %}. I put the overriden file change_form.html in pricingmeister/accounts/templates/admin/. But i could not see the change. The hierarchy of my Django Project and folders is below: Django Project Hierarchy Screenshot Below is my settings.py (truncated some code to show only relevant part of code . . . INSTALLED_APPS = [ # … -
List View is not working but get context data is
I have a ListView but when I call it only the get_context_data method works (the news and category model, not the product) when I try to display the information of the models in the templates. view: class HomeView(ListView): model = Product context_object_name='products' template_name = 'main/home.html' paginate_by = 25 def get_context_data(self, **kwargs): categories = Category.objects.all() news = News.objects.all() context = { 'categories' : categories, 'news' : news, } context = super().get_context_data(**kwargs) return context There is also this piece of code: context = super().get_context_data(**kwargs) If it's written before: categories = Category.objects.all() The Product model is show but not the others. base.html <body> ... {% include "base/categories.html" %} {% block content %}{% endblock %} </body> home.html {% extends 'main/base.html' %} {% block content %} <div> ... <div> {% for product in products %} {% if product.featured == True %} <div> <div> <a href="">{{ product.author }}</a> <small>{{ product.date_posted|date:"F d, Y" }}</small> </div> <p>Some text..</p> </div> {% endif %} {% endfor %} </div> </div> {% endblock content %} categories.html <div> ... <div> {% for category in categories %} <p>{{ category.name }}</p> {% endfor %} </div> <div> {% for new in news %} <p>{{ new.title }}</p> {% endfor %} </div> </div> -
How to delete 200,000 records with DJango?
Situation: I have Model have a relation 1-1, sample: class User(models.Model): user_namme = models.CharField(max_length=40) type = models.CharField(max_length=255) created_at = models.DatetimeField() ... class Book(models.Model): user = models.OneToOneField(User, on_delete=models.CASCADE) And I have a around 200,000 records. Languague: Python Framework: Django Database: Postgres Question: How can I delete 200,000 records above with minimal cost? Solution I have tried: user_ids = Users.objects.filter(type='sample', created_date__gte='2022-11-15 08:00', created_date__lt="2022-11-15 08:30").values_list('id',flat=True)[:200000] # Fetch 200,000 user ids. for i, _ in enumerate(user_ids[:: 1000]): with transaction.atomic(): batch_start = i * self.batch_size batch_end = batch_start + self.batch_size _, deleted = Users.objects.filter(id__in=user_ids[batch_start,batch_end] With this solution, my server use arround: 600MB CPU 300MB RAM Take more 15 minutes to finish workload. I wonder do anyone have a better solution? -
How to ignore errors in twilio send sms
I am trying to send sms using python and twilio. I am actually sending 1000-2000 sms. so while sending I get errors like : 30032 - Toll-Free Number Has Not Been Verified 21610 - Attempt to send to unsubscribed recipient so here what I am trying. I have a form inside my dashboard. When I will press send button it will start sending message to every one. when someone blocks my number from there side the whole process get closed and I had to send messages or remove them from my list again and again. Here is views.py class AdminIndex(View): template_name = "AdminPage/index.html" def get(self, request): contents = ContentTemplate.objects.all() numbers = NumberGroup.objects.all() groups = Group.objects.all() args = { "contents": contents, "numbers": numbers, "groups": groups } return render(request, self.template_name, args) def post(self, request): content = request.POST.get("content") number = request.POST.get("number") groups = request.POST.get("group") if request.method == "POST": message = SendMessageModel( content=ContentTemplate.objects.get(id=content), number=NumberGroup.objects.get(id=number), group=Group.objects.get(id=groups) ) message.save() for n in message.group.customers.all(): account_sid = "AC2b0cc7c783ccc1e82f3771636dda5e73" auth_token = "bdb32f3656485e868270f68a1b3024ee" client = Client( account_sid, auth_token ) send_message = client.messages.create( body=message.content.content, from_=f"+{message.number.number}", to=n.phone_number, ) n.is_sent = True print(str(n.phone_number)) n.save() return HttpResponse("All Sent") So, How can I ignore these errors and keep sending messages to the users without interruption?
