Django community: RSS
This page, updated regularly, aggregates Django Q&A from the Django community.
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how to use msg91 in django?
def send_email(request): conn = http.client.HTTPSConnection("api.msg91.com") payload = "{\r\n\"to\": [\r\n{\r\n\"name\": \"Test\",\r\n\"email\": \"email\"\r\n}\r\n],\r\n\"from\": {\r\n\"name\": \"Joe\",\r\n\"email\": \"email\"\r\n},\r\n\"domain\": \"my domian name\",\r\n\"mail_type_id\": \"1\",\r\n\"template_id\": \"mytemplate-name\",\r\n\"variables\": {\r\n\"VAR1\": \"12345\",\r\n\"VAR2\": \"1234\"\r\n},\r\n\"authkey\": \"myauthkey\"\r\n}" headers = { 'Content-Type': "application/JSON", 'Accept': "application/json" } conn.request("POST","/api/v5/email/send",payload,headers) res = conn.getresponse() data = res.read() print(data.decode("utf-8")) return None -
Showing Operation object (1) when trying to get foreignkey django
I am trying to post a table in HTML and i get This "Operation object (1)" insted of just the id "1" How is posible to fix this as you can see in the picture where is selected i want to have the id not text models.py operationID = models.IntegerField(primary_key=True) assignee = models.ForeignKey('Employee', on_delete=models.CASCADE) dateRegistered = models.DateTimeField(auto_now_add=True) timeStart = models.DateTimeField(auto_now_add=True, null=True) timeFinish = models.DateTimeField(null=True) status = models.ForeignKey( 'Status', on_delete=models.CASCADE) class Subtask(models.Model): subtaskID = models.IntegerField(primary_key=True, auto_created=True) operationID = models.ForeignKey('Operation', on_delete=models.CASCADE) containerID = models.CharField(max_length=255) containerWeightT = models.DecimalField(max_digits=6, decimal_places=2) loadSeq = models.IntegerField() moveTo = models.ForeignKey('MoveTo', on_delete=models.CASCADE) stow = models.ForeignKey('Stow', on_delete=models.CASCADE) status = models.ForeignKey( 'Status', on_delete=models.CASCADE)``` views.py ```def displaydata(request): results1 = Subtask.objects.prefetch_related( 'moveTo', 'operationID', 'stow', 'status').all() return render(request, 'ee.html', {'Subtask': results1})``` [website][1] [1]: https://i.stack.imgur.com/cWFdd.png -
remove div only on one page in Django template
I have a website that contains 15 seperate pages. On each page I have a div named header. Right know I am removing this element via JavaScript like this: homepage.html <script> const element = document.getElementById('header'); element.remove(); </script> The problem is that I can see the header element for a second before it is removed. The script is placed on the top of the page. I also tried to do it via css like this: .header { height: 0rem !important; visibility: hidden !important; } However, I still see the element disappearing when refreshing the page. I know that I could create a template block in my base.html file and just exclude it in the homepage.html page but I'd need to include it in the other 14 HTML files manually. Question Is there any other/better way to exclude div so I don't see it at all on my homepage? -
Django, django-filter and pagination
my goal is to have a 'user_profile' page that displays relevant information of the user of interest. Furthermore, the 'user_profile' page should include all the posts that were created by the respective user as new blog entries. These posts, however, should be filterable with the application 'django-filter' and be paginated. At the moment I have difficulties to paginate the filtered posts. So my question is how to achieve the latter? So far, I used following approach: filters.py import django_filters class AccountPostFilter(django_filters.FilterSet): title = django_filters.CharFilter(lookup_expr='icontains') category = django_filters.ChoiceFilter(choices=cat_list) class Meta: model = Post fields = ['title', 'category'] views.py class UserProfile(DetailView, MultipleObjectMixin): model = Account template_name = 'account/user_profile.html' paginate_by = 5 def get_context_data(self, **kwargs): posts = Post.objects.all().filter(author=self.kwargs['pk']) context = super().get_context_data(object_list=posts, **kwargs) context['filterset'] = AccountPostFilter(self.request.GET, queryset=posts) return context Thank you very much for your time. Best wishes, Daniel -
Django ElasticBeanstalk Deploy- error deterministic=True requires SQLite 3.8.3 or higher
so, i deployed my application in my local venv i did python make migrations python migrate and i did eb deploy and eb status the helth returned green so its working, but when i enter the web site it returns deterministic=True requires SQLite 3.8.3 or higher Note: Locally it works just fine commands that i ran to make my project: python manage.py mamemigrations python manage.py migrate python manage.py createsuperuser eb init python-3.8 Naameofmyproject eb create Nameofmyproject Requirments.txt: asgiref==3.5.0 autopep8==1.6.0 certifi==2021.10.8 charset-normalizer==2.0.12 dj-database-url==0.5.0 Django==4.0.3 django-anymail==8.5 django-autoslug==1.9.8 django-crispy-forms==1.14.0 django-environ==0.8.1 django-model-utils==4.2.0 idna==3.3 Pillow==9.1.0 psycopg2-binary==2.9.3 pycodestyle==2.8.0 python-dateutil==1.5 requests==2.27.1 six==1.16.0 sqlparse==0.4.2 stripe==2.70.0 toml==0.10.2 tzdata==2022.1 urllib3==1.26.9 settings.py: DATABASES = { 'default': { 'ENGINE': 'django.db.backends.sqlite3', 'NAME': BASE_DIR / 'db.sqlite3', } } STATIC_URL = '/static/' STATIC_ROOT = BASE_DIR / 'static' STATICFILES_DIRS = [BASE_DIR / 'templates/static'] MEDIA_URL = '/media/' MEDIA_ROOT = BASE_DIR / 'media' CART_SESSION_ID = 'cart' AUTH_USER_MODEL = 'account.UserBase' LOGIN_REDIRECT_URL = '/account/dashboard' LOGIN_URL = '/account/login/' EMAIL_BACKEND = 'django.core.mail.backends.console.EmailBackend' DEFAULT_AUTO_FIELD = 'django.db.models.BigAutoField' -
numpy heroku error: legacy-install-failure
I'm trying to deploy django webapp on heroku but getting error: legacy-install-failure on numpy package. Building on Heroku-20 stack, default python-3.10.4, pip 22.0.4. note: This is an issue with the package mentioned above, not pip -
OSX monterey -> ld: library not found for -lintl
When pipenv install uwsgi This error comes, ld: library not found for -lintl I guess it requires X-code library ? I tried this as well $xcode-select --install xcode-select: error: command line tools are already installed, use "Software Update" to install updates However it is already installed. How can I solve this? -
I tried to add sub frontend project in my templates but I got error "MIME checking is enabled."?
goal I have two frontend project and I want to do create a new folder called build_2 inside my templates folder. inside the build_2 have index.html file and it was working just fine problem but where I added <link href="my_styles.css" rel="stylesheet"> I got this error Note: the directory myProject/home/templates/static/css/my_styles.css full error message: Refused to apply style from 'http://127.0.0.1:8000/css/my_styles.css' because its MIME type ('text/html') is not a supported stylesheet MIME type, and strict MIME checking is enabled. I tried add #settings.py TEMPLATES['DIRS'].append(os.path.join(BASE_DIR, 'home/templates/build2'), os.path.join(BASE_DIR, 'home/templates')) add STATICFILES_DIRS.append(os.path.join(BASE_DIR, 'home/templates/static')) but I am still getting the same error? -
download and add files in vuejs and django
how can i add a file in my front end vuejs and download my file from my backend django rest framework with v-model in vuejs my modele is article , class Article(models.Model): conference = models.ForeignKey(Conference,related_name='articles',on_delete=models.CASCADE) title =models.CharField(max_length=50) Auteur =models.CharField(max_length=50) resume =models.TextField() motcles =models.TextField() create_by = models.ForeignKey(User, related_name='articles',on_delete=models.CASCADE) fichier = models.FileField(upload_to='uploads',blank=True, null=True) def __str__(self): return self.title this code is about how i add a new article , i should here add how to enter a file <form v-on:submit.prevent="submitArticle()"> <div class="field"> <label class="label">Title</label> <div class="control"> <input type="text" class="input" v-model="article.title"> </div> </div> <div class="field"> <label>Auteur</label> <div class="control"> <input type="title" class="input" v-model="article.Auteur"> </div> </div> <div class="field"> <label class="label">Resume</label> <div class="control"> <textarea type="text" class="input" v-model="article.resume"></textarea> </div> </div> <div class="field"> <label>motcles</label> <div class="control"> <input type="title" class="input" v-model="article.motcles"> </div> </div> <div id="app"> <a href="#" @click.prevent=" downloadItem({ url: 'https://test.cors.workers.dev/?https://www.w3.org/WAI/ER/tests/xhtml/testfiles/resources/pdf/dummy.pdf', label: 'example.pdf', }) " > download </a> </div> <div class="field"> <div class="control"> <button class="button is-link">submit</button> </div> </div> </form> this is how i enter my article submitArticle() { console.log('submitArticle') const conferenceID = this.$route.params.id this.errors = [] if (this.article.title === '') { this.errors.push('The title must be filled out') } if (this.article.resume === '') { this.errors.push('The content must be filled out') } if (!this.errors.length) { axios .post(`/api/v1/add-article/${conferenceID}/`, this.article) .then(response => { this.article.title … -
Django - Permissions in file upload
I'm trying to build a file sharing application using Django REST Framework in the backend. In order to upload files, I have the following field in my models: user_file = models.FileField() However, when a user uploads a file, that file is stored in the directory of the DRF project, and if I put the path of that file in a browser, every user can then access it as well. How can I prevent this? I've thought of having an Apache HHTPd server where there would be a folder for each user and when a user tries to access a file that was not uploaded my them, the backend would do that verification and would not allow it, but I don't know if that would solves the problem? How can I do this? Thanks -
Django - Show all ForeignKeys belonging to the given ID as a select multiple field
I have this model: class OrderProduct(models.Model): order = models.ForeignKey(to=Order, on_delete=models.CASCADE) product = models.ForeignKey(to=Product, on_delete=models.CASCADE) quantity = models.PositiveIntegerField(default=1) price_paid = models.DecimalField(max_digits=5, decimal_places=2) @property def total_value(self): return self.price_paid * self.quantity def __str__(self): return f"{self.order.id} / {self.order.user} // {self.product.name} / {self.quantity} ks" def save(self, *args, **kwargs): self.price_paid = self.product.price super(OrderProduct, self).save(*args, **kwargs) This is my form: class ChangeOrderProductForm(ModelForm): class Meta: model = OrderProduct fields = ('product',) This is my view: def change_orderproduct(request, order_id: int): set_session_cookie_restraunt_status(request) if not check_if_user_has_permission(request, 'kitchen.change_orderproduct'): messages.warning(request, 'Unauthorized to perform this action.') return redirect("index") ordered_products = OrderProduct.objects.filter(order__pk=order_id).first() order = Order.objects.filter(pk=order_id).first() if not order: messages.info(request, "Given order does not exist.") return redirect(request.META.get('HTTP_REFERER', index)) if request.method == "GET": form = ChangeOrderProductForm() context = { "form": form, "order": order } return render(request, "kitchen/change_orderproduct.html", context=context) if request.method == "POST": form = ChangeOrderProductForm(data=request.POST, instance=ordered_products) if form.is_valid(): form.save() messages.success(request, f"Products in order {order.pk} were successfully changed.") return redirect(request.META.get('HTTP_REFERER', all_orders_view)) The code above works, but it only allows you to choose 1 Product in the ChangeOrderProductForm view: However, there are multiple products for almost all Order instances. How do I make it to: Show all the possible Product instances to choose from in the form? Make the already existing ForeignKeys assigned to OrderProduct as the pre-selected ones? … -
find top performing list of category which have highest number of orders in django
models.py class Line_items(models.Model): id = models.AutoField(primary_key=True) product = models.ForeignKey('Products' , on_delete=models.DO_NOTHING ) class Products(models.Model): id = models.AutoField(primary_key=True) name = models.CharField(max_length=400 , blank=True) category = models.ManyToManyField('Categories', through='Product_Categories', related_name='products') class Categories(models.Model): id = models.AutoField(primary_key=True) name = models.CharField(max_length=100 , blank=True) slug = models.CharField(max_length=200 , blank=True) class Product_Categories(models.Model): id = models.AutoField(primary_key=True) product_id = models.ForeignKey(Products, on_delete=models.DO_NOTHING) category_id = models.ForeignKey(Categories, on_delete=models.DO_NOTHING) here are my models. where line_items contains number of orders done till now. in line_items we have connect product id with product table. but we don't have any connetion from product table to category table. ( category table contains every category and their id ). to connect product table with category table we have created new table 'product_categories' which connects each category with their respective product. here what we want is top performing category. category which have highest number of orders. thanks -
TypeError: hasattr(): attribute name must be string (Django)
Hi how to solve this issue, when i am excluding this code i am getting this type error. my admin.py is this, list_display = [ "id", exclude==("mode_of_action",)] -
AWS Lambda, An error occurred (InvalidToken) when calling the PutObject operation: The provided token is malformed or otherwise invalid
I made a Django application and tried to deploy it using Zappa and AWS Lambda. My deployment was successfully deployed but, the image was not uploaded to AWS S3 to invoke the API. This is my cloudwatch error log [ERROR] 2022-04-22T08:35:19.84Z cbf18c70-f478-4363-8f5a-0777c76564e9 Internal Server Error: /production/v1/ReviewCamping/ Traceback (most recent call last): File "/var/task/django/core/handlers/exception.py", line 55, in inner response = get_response(request) File "/var/task/django/core/handlers/base.py", line 197, in _get_response response = wrapped_callback(request, *callback_args, **callback_kwargs) File "/var/task/django/views/decorators/csrf.py", line 54, in wrapped_view return view_func(*args, **kwargs) File "/var/task/rest_framework/viewsets.py", line 125, in view return self.dispatch(request, *args, **kwargs) File "/var/task/rest_framework/views.py", line 509, in dispatch response = self.handle_exception(exc) File "/var/task/rest_framework/views.py", line 469, in handle_exception self.raise_uncaught_exception(exc) File "/var/task/rest_framework/views.py", line 480, in raise_uncaught_exception raise exc File "/var/task/rest_framework/views.py", line 506, in dispatch response = handler(request, *args, **kwargs) File "/var/task/rest_framework/mixins.py", line 19, in create self.perform_create(serializer) File "/var/task/service/views/v1/camping.py", line 80, in perform_create serializer.save(owner=self.request.user) File "/var/task/rest_framework/serializers.py", line 212, in save self.instance = self.create(validated_data) File "/var/task/rest_framework/serializers.py", line 962, in create instance = ModelClass._default_manager.create(**validated_data) File "/var/task/django/db/models/manager.py", line 85, in manager_method return getattr(self.get_queryset(), name)(*args, **kwargs) File "/var/task/django/db/models/query.py", line 514, in create obj.save(force_insert=True, using=self.db) File "/var/task/django/db/models/base.py", line 806, in save self.save_base( File "/var/task/django/db/models/base.py", line 857, in save_base updated = self._save_table( File "/var/task/django/db/models/base.py", line 1000, in _save_table results = self._do_insert( … -
Is it possible to link multiple models to one fiel in django?
Let's say I have these models: class Material(models.Model): name = models.CharField([...]) class Consumable(models.Model): name = models.CharField([...]) restores = models.IntegerField([...]) class Weapon(models.Model): name = models.CharField([...]) damage = models.IntegerField([...]) # And then I have an 'inventory', like this one: class Inventory(models.Model): user = models.ForeignKey([...]) # to which user you want to link the item item = models.ForeignKey([...]]) # which item quantity = models.IntegerField([...]) # how many of it I want to be able to have all Material, Consumable, and Weapon models listed in the 'item' field, so when you want to add an item as an inline, you would see all 3 models' objects. Something like # instead of this item = models.ForeignKey(Consumable) # which item # want something like this item = models.ForeignKey(Consumable and Material and Weapon) # which item # this wouldn't work ofc... Is there a way to collect all 3 of them and pass them to the 'item' field, without the need of restarting the server? (when making a "choices" list that queries from a model you must restart the server to see the newly added objects, I don't want that.) I also want to stick to the built-in admin of Django since it provided everything I need … -
How to add similar key value in python?
Sorry if you find the question misleading This is the array of data: [{'user': 1, 'coins': 2}, {'user': 18, 'coins': 8}, {'user': 1, 'coins': 1}, {'user': 3, 'coins': 1}, {'user': 5, 'coins': 1}, {'user': 7, 'coins': 0}, {'user': 18, 'coins': 0}] Expected Outcome: [{'user': 1, 'coins': 3}, {'user': 18, 'coins': 8}, {'user': 3, 'coins': 1}, {'user': 5, 'coins': 1}, {'user': 7, 'coins': 0}] So, there will be an array of those data, and we have to add the coins to similar user which can be identified by user key value. If user key have similar value that is 1 (in above case) then coins of both or multiple user must be added making it a total sum of that user coins. -
Password validation with CustomUser Django
I am trying to validate a password against CustomUser fields: email and full_name. All test validations are working except for UserAttributeSimilarityValidator, which is the only test that I have included with the code below. forms.py class RegistrationForm(forms.ModelForm): email = forms.EmailField(label=_('Email address'), widget=forms.EmailInput(attrs={'class': 'form-control mb-4', 'class': 'form-control mb-4', })) full_name = forms.CharField(label=_('Full name'), widget=forms.TextInput(attrs={'class': 'form-control mb-4', })) password = forms.CharField(label=_('Password'), widget=forms.PasswordInput(attrs={'class': 'form-control mb-4', 'autocomplete': 'new-password', 'id': 'psw', })) class Meta: model = get_user_model() fields = ('full_name', 'email', 'password') def clean_password(self): password = self.cleaned_data.get("password") if password: try: password_validation.validate_password(password, self.instance) except ValidationError as error: self.add_error("password", error) return password def clean_email(self): email = self.cleaned_data['email'] if User.objects.filter(email=email).exists(): raise forms.ValidationError( mark_safe(_(f'A user with that email already exists, click this <br><a href="{reverse("account:pwdreset")}">Password Reset</a> link' ' to recover your account.')) ) return email tests.py @override_settings( AUTH_PASSWORD_VALIDATORS=[ { 'NAME': 'django.contrib.auth.password_validation.UserAttributeSimilarityValidator', 'OPTIONS': { 'user_attributes': ( 'email', 'full_name', )}, }, ] ) def test_validates_password(self): data = { "email": "jsmith@example.com", "full_name": "John Smith", "password": "jsmith", } form = RegistrationForm(data) self.assertFalse(form.is_valid() self.assertEqual(len(form["password"].errors), 1) self.assertIn( "The password is too similar to the email.", form["password"].errors, ) Which results in this test result: FAIL: test_validates_password (account.tests.test_forms.RegistrationFormTest) ---------------------------------------------------------------------- Traceback (most recent call last): File "/Users/tester/Documents/dev/test/venv/lib/python3.8/site-packages/django/test/utils.py", line 437, in inner return func(*args, **kwargs) File "/Users/tester/Documents/dev/test/account/tests/test_forms.py", line 112, in … -
How to make files available not in "/static/", but in "/app/static/"?
I have Django 1.11 project structure like this: project static img1.png app static img2.png File "settings.py" contains: STATIC_URL = '/static/' STATICFILES_DIRS = [ os.path.join(BASE_DIR, 'static'), ] Images are available at: my-site.com/static/img1.png my-site.com/static/img2.png But I want the images to be available in a different way: my-site.com/app/static/img1.png my-site.com/app/static/img2.png I can write to a file "project/app/urls.py": urlpatterns += static('static/', document_root=os.path.join(BASE_DIR, 'app' + STATIC_URL)) And then "img2.png" will be available, but not "img1.png". How can I make "img1.png" available? -
how to create row in model based on data fetched from JSON request from third-party crawling with "best practice"
describe the problem: I want to crawl data from dataforseo , and save them directly in model database through model.create() method with having multi model with multi relation with models so for instance in model A have ManyToMany relation with model B ManyToMany relation with model C ManyToMany relation with model D and model B have relation with model C so my question is how to save JSON response to all model mentioned above smoothly through model A create Method Code: view.py file @api_view(['POST']) @parser_classes((JSONParser,)) def crawl_data(request): """ A view that can accept POST requests with JSON content. """ Product.create( title=request.data[title] url=request.data[url] describtion=request.data[describtion] ... ) return Response({'received data': request.data}) models.py class Highlighted(models.Model): name_highlighted = models.CharField(max_length=100) def __str__(self): return str(self.name_highlighted) class Rating(models.Model): rating_type = models.CharField(max_length=500, null=True, blank=True) # make unique value = models.CharField(max_length=500, null=True, blank=True) votes_count = models.CharField(max_length=500, null=True, blank=True) rating_max = models.CharField(max_length=500, null=True, blank=True) def __str__(self): return str(self.value) class Price(models.Model): current = models.CharField(max_length=500, null=True, blank=True, default="none") regular = models.CharField(max_length=500, null=True, blank=True) max_value = models.CharField(max_length=500, null=True, blank=True) def __str__(self): return str(self.current) class Product(models.Model): title = models.CharField(max_length=500, null=True, blank=True) url = models.CharField(max_length=500, null=True, blank=True) description = models.CharField(max_length=500, null=True, blank=True) pre_snippet = models.CharField(max_length=500, null=True, blank=True) extended_snippet = models.CharField(max_length=500, null=True, blank=True) images = models.CharField(max_length=500, … -
Django Admin dependent DropDown HTML field
I have a Region and SubRegion ForeignKey in the Country Model. My SubRegion model also has a Region ForeignKey for the Region Model. I am having a hard time displaying the SubRegion dropdown in the Country Model on the basis of the Region selected. What would be the best way to achieve it? It would be great if dropdowns were based on the value of their parent. models.py class Region(models.Model): name = models.CharField(max_length=20) is_active = models.BooleanField(verbose_name='is active?', default=True) created_at = models.DateTimeField(auto_now_add=True) updated_at = models.DateTimeField(auto_now=True) def __str__(self): return self.name class SubRegion(models.Model): region = models.ForeignKey(Region, on_delete=models.CASCADE) name = models.CharField(max_length=50) is_active = models.BooleanField(verbose_name='is active?', default=True) created_at = models.DateTimeField(auto_now_add=True) updated_at = models.DateTimeField(auto_now=True) def __str__(self): return self.name class Meta: verbose_name_plural = 'Sub Regions' class Country(models.Model): name = models.CharField(max_length=100) region = models.ForeignKey(Region, on_delete=models.CASCADE, verbose_name='Region') subregion = models.ForeignKey(SubRegion, on_delete=models.CASCADE, verbose_name='Sub Region') created_at = models.DateTimeField(auto_now_add=True) updated_at = models.DateTimeField(auto_now=True) admin.py class CountryAdmin(admin.ModelAdmin): list_display = ('name', 'is_active', 'is_default', 'created_at', 'updated_at', ) list_filter = ('region', 'subregion', 'is_active', 'is_default', 'created_at', 'updated_at', ) search_fields = ('name', 'official_name', ) fieldsets = ( ('Region Information', {'fields': ('region', 'subregion', )}), ('Basic Information', {'fields': ('name', 'official_name', )}), ('Codes Information', {'fields': ('cca2', 'ccn3', 'cca3', 'cioc','idd', 'status', )}), ('Coordinates', {'fields': ('latitude', 'longitude', )}), ('Membership & Statuses', {'fields': (('is_independent', 'is_un_member', … -
Django - if i add multiple row in views, its not showing up before i refresh PostgreSQL services
for E in PositionList.objects.all(): # It fetching above 100 records AE = Attendance() AE.EmployeeName = Candidates.objects.get(pk=E.EmployeeName_id) AE.Company = Preclient.objects.get(pk=E.Company_id) AE.AttendanceDate = date.today() AE.Attendance = True AE.Remarks = "Initial Entry" AE.save() It actually created above 100 records, but it wont showing up. if refresh or restart all services like PostgreSQL, it will showing up sudo systemctl restart gunicorn sudo systemctl daemon-reload sudo systemctl restart gunicorn.socket gunicorn.service sudo nginx -t && sudo systemctl restart nginx **if i do following commands in my console, it will be showing up What is the solution for it??? i am using gunicorn django nginx ubuntu** -
Django complex multiform views
I am not that good in coding, so take me easy. I would like to create/update view with multiple models in a single form , but need to dynamically add second and third model (need formsets) form into the view. That could be achieved by javascript , by copy empty forms. here is model structure, need it to be scalable, but lets stack with 3 models. What I need 1)present parent and second 2)hit resolve button , which will send ajax and all form data then I proces validate and get additional data and that generate third forms 3)save everything back to database , with appropriate ForeigKeys between models models.py from django.db import models class Parrent(models.Model): data = models.CharField(max_length=30) class Second(models.Model): parent = models.ForeignKey(Parrent, on_delete=models.CASCADE) data = models.CharField(max_length=30) class Third(models.Model): second = models.ForeignKey(Second, on_delete=models.CASCADE) data = models.CharField(max_length=30) forms.py from django.forms import ModelForm from myapp.models import Parrent, Second, Third class ParrentForm(ModelForm): class Meta: model = Parrent fields = ['data'] class SecondForm(ModelForm): class Meta: model = Second fields = ['data'] class ThirdForm(ModelForm): class Meta: model = Third fields = ['data'] SecondFormSet = formset_factory(SecondForm, extra=1) ThirdFormSet = formset_factory(ThirdForm, extra=1) template.html <!-- [ Template form second] start --> <div id='template-second-form' class="card-body table-border-style hidden"> {{ … -
Filter queryset by checking two columns if they are equal
models.py class Entry(models.Model): paid = models.FloatField(blank=True, null=True) price = models.FloatField(blank=True, null=True) I want to get all entries that have paid and price columns the same value. How can I achieve this? -
INSERT failed: NOT NULL constraint failed: error in django and sqlite3
when i run this command in sqlite3: sqlite> .import af.csv tmpr_product it gives me this error: af.csv:1: expected 2 columns but found 1 - filling the rest with NULL af.csv:1: INSERT failed: datatype mismatch af.csv:2: expected 2 columns but found 1 - filling the rest with NULL af.csv:2: INSERT failed: NOT NULL constraint failed: tmpr_product.number af.csv:3: expected 2 columns but found 1 - filling the rest with NULL af.csv:3: INSERT failed: NOT NULL constraint failed: tmpr_product.number af.csv:4: expected 2 columns but found 1 - filling the rest with NULL af.csv:4: INSERT failed: NOT NULL constraint failed: tmpr_product.number af.csv:5: expected 2 columns but found 1 - filling the rest with NULL af.csv:5: INSERT failed: NOT NULL constraint failed: tmpr_product.number af.csv:6: expected 2 columns but found 1 - filling the rest with NULL af.csv:6: INSERT failed: NOT NULL constraint failed: tmpr_product.number af.csv:7: expected 2 columns but found 1 - filling the rest with NULL af.csv:7: INSERT failed: NOT NULL constraint failed: tmpr_product.number af.csv:8: expected 2 columns but found 1 - filling the rest with NULL af.csv:8: INSERT failed: NOT NULL constraint failed: tmpr_product.number af.csv:9: expected 2 columns but found 1 - filling the rest with NULL af.csv:9: INSERT failed: NOT NULL constraint … -
Run django management command for admin file
I want to run this management command but its not working for me , from django.core.management.base import BaseCommand from product.admin import ProductinfoAdmin class Command(BaseCommand): def handle(self, *args, **options): moa = ProductinfoAdmin.admin_site.filter(list_display("info")).update(list_display=False) print(f"The total number of activated Products {moa}")
