Django community: RSS
This page, updated regularly, aggregates Django Q&A from the Django community.
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Django form is not showing is not generating in html template
I am using django.contrib.auth.views.LoginView for views and here are my files html <form method="GET"> {% csrf_token %} {{ formlog|crispy }} </form> views.py def index(request): formlog = auth_views.LoginView.as_view() if request.method == 'POST': form = UserRegisterForm(request.POST) if form.is_valid(): form.supervalid() form.save() username = form.cleaned_data.get('username') messages.success(request, f'Dear {username} you have been created a new accound!') return redirect('main') else: form = UserRegisterForm() return render(request, 'main/index.html', {'form': form, 'formlog': formlog}) forms.py from django import forms from django.contrib.auth.models import User from django.contrib.auth.forms import UserCreationForm class UserRegisterForm(UserCreationForm): email = forms.EmailField() name_and_surname = forms.CharField(min_length=5, max_length=30) def supervalid(self): expr_a = User.objects.filter(name_and_surname=self.cleaned_data['name_and_surname']).exists() expr_b = User.objects.filter(email=self.cleaned_data['email']).exists() if expr_b: raise ValidatioError(f'There already exists user with that email, use another one ;)') if expr_a: raise ValidatioError(f'This name is already used, sorry') class Meta: model = User fields = ['username', 'email', 'password1', 'password2'] please help istead of talking about things that dont metter. Thanks for help ;) -
Django doesn't load static file and throws error although syntax is correct - is VS Code the culprit?
I have been working with Django for a few days and I'm trying to get the hang of static files and how to integrate them in my project. I have the following code: <div class="background_image" style="background-image: url({% static 'images/home_slider.jpg' %});"> </div> And Python throws this error for this specific block of code: I have more than 30 cases in the same file where I used this pattern {% static 'something' %} and it works in all cases. The only difference between these cases in the one above (that's not working) is that the one above is the only one where I apply the style directly on the HTML tag. I am am my wits' end. From what I researched, the syntax seems to be correct for adding a background image directly in HTML so I'm starting to believe that the error is caused by the fact that VS code is formatting my code in a super weird way when saving and I get something like: <div class="background_image" style=" background-image: url({%static'images/home_slider.jpg'%}); " ></div> Any suggestion would be appreciated. Thank you. -
ValueError at /signup/ The given username must be set
so basically I have been building this blogging website and now I am stuck at this point of the Signup process, the whole work is done using django views.py: def handleSignup(request): if request.method == 'POST': # getting user parameters username = request.POST.get('username') fname = request.POST.get('fname') lname = request.POST.get('lname') email = request.POST.get('email') pass1 = request.POST.get('pass1') pass2 = request.POST.get('pass2') # fname = request.POST['fname'] # lname = request.POST['lname'] # email = request.POST['email'] # pass1 = request.POST['pass1'] # pass2 = request.POST['pass2'] # creating users myuser = User.objects.create_user(username = username, email = email, password = pass1) myuser.first_name = fname myuser.last_name = lname myuser.save() messages.success(request, 'your account have been successfully created!') return redirect(request, 'home') else: return HttpResponse("error 404 not found") form in base.html: <form action="/signup/" method="post"> <div class="form-group"> <label for="username">Username</label> <input type="text" class="form-control" id="username" placeholder="choose a unique username"> </div> <div class="form-group"> <label for="fname">Firstname</label> <input type="text" class="form-control" id="fname" placeholder="First Name"> </div> <div class="form-group"> <label for="lname">Lastname</label> <input type="text" class="form-control" id="lname" placeholder="Last Name"> </div> <div class="form-group"> <label for="email">Email</label> <input type="email" class="form-control" id="email" placeholder="email@example.com"> </div> <div class="form-group"> <label for="pass1">Choose Password</label> <input type="password" class="form-control" id="pass1"> </div> <div class="form-group"> <label for="pass2">Confirm password</label> <input type="password" class="form-control" id="pass2"> </div> {% csrf_token %} <button type="submit" class="btn btn-primary">Submit</button> </form> now I am getting this error: ValueError … -
Django path in admin absolute instead relative
In my admin project i have a situation like this: settings.py: STATIC_URL = '/myproject/static/' models.py: class v_candidatura(models.Model): ... c_cv = models.FileField(upload_to='myproject/static/docs', max_length=254, validators=[FileExtensionValidator(['pdf']),validate_fsize], verbose_name="CV") .... admin.py: class vcandidaturaAdmin(admin.ModelAdmin): list_filter = ('c_mansione',) list_display = ('c_data', 'c_sur', 'c_name', 'c_dob', 'c_en', 'c_de', 'c_cv') ordering = ('-c_data',) admin.site.register(v_candidatura, vcandidaturaAdmin) Now if i log into admin site and go to the 127.0.0.1:8000/admin/myapp/v_candidatura/ and click on c_cv link Django atach begore c_cv link the wole path 127.0.0.1:8000/admin/myapp/v_candidatura/, how can i define my href link for ahref in django admin site? for example i would clicking on my c_cv django admin list field open 127.0.0.1:8000/<c_cv path> So many thanks in advance admin.py -
django ecommerce communication
I am building an e-commerce app for fast food company using Django. There will be a dash-board based on Django admin that will give the vendor control over activating offers or closing shop etc. What I am trying to implement is that once the customer places the order and a message flashes for the vendor to accept or reject. I am not sure what I will need to use to implement that two-way communication between the customer and the vendor admin. Any ideas or alternatives will be very helpful. Thanks -
How to setup carousel slider in Django with records?
I am trying to implement carousel slider in my website, I have ForeignKey relation between Project and Image, I have multiple image store in Image model. It Means that a project can have multiple images, and i want to display those project images in slider, Please let me know how I can display the images in slider, Because currently my images are coming in vertical format. Here is my index.html code, here i am trying to display all images in slider, but it's not working, Please check my code and solve my issue. <div id="carousel-thumb" class="carousel slide carousel-fade carousel-thumbnails list-gallery pt-2" data-ride="carousel"> <!--Slides--> <div class="carousel-inner" role="listbox"> {% for i in project.projectimage.all %} <div class="carousel-item active"> <img class="d-block w-100" src="{{i.project_image.url}}" alt="slide" style="height: 420px;"> </div> {% endfor %} </div> <!--Controls starts--> <a class="carousel-control-prev" href="#carousel-thumb" role="button" data-slide="prev"> <span class="lnr lnr-arrow-left" aria-hidden="true"></span> <span class="sr-only">Previous</span> </a> <a class="carousel-control-next" href="#carousel-thumb" role="button" data-slide="next"> <span class="lnr lnr-arrow-right" aria-hidden="true"></span> <span class="sr-only">Next</span> </a> </div> -
django rest framework defaultRouter without models
I am new to Python and Django (rest framework) and I love the DefaultRouter feature, by also giving an "API ROOT" page. But, I can only use the defaultrouter.register() method with viewsets. How to make a custom route without a model? For example I will have a "POST" route that multiplies 2 values from the body. (so body is something like {"value1":123, "value2":456} Is it good practice to use viewsets for everything? How should I implement a custom (multiply) function? I will have a serializer with the 2 params? And where to implement the code? Am I doing things right? -
Load data in memory during server startup in Django app
I am creating a Django app wherein I need to load and keep some data in memory when the server starts for quick access. To achieve this I am using Django's AppConfig class. The code looks something like this : from django.core.cache import cache class myAppConfig(AppConfig): name = 'myapp' def ready(self): data = myModel.objects.values('A','B','C') cache.set('mykey', data) The problem is that this data in cache will expire after sometime. One alternative is to increase the TIMEOUT value. But I want this to be available in memory all the time. Is there some other configuration or approach which I can use to achieve this ? -
Using Q object with variable
I'd like to use the django.db.models.Q object in a way that the query term is coming from a variable. What i'd like to achieve is identical to this: q = Q(some_field__icontains='sth') Obj.objects.filter(q) , but the some_field value should come from a variable: field_name='some_field' q = Q('%s__icontains=sth' % field_name) Obj.objects.filter(q) , but this solution does not give me the correct result of course. I also tried to use dictionary this way: dt = {'%s__icontains' % field_name: 'sth'} q = Q(**dt) Obj.objects.filter(q) , but this also fails on the result. How could I use the Q object using variables as query term? Thanks. -
Django edit form won't validate on form.save() with user field
I recently added a "user" field to my Game model. I can create a new game that works fine; it's when I want to allow a user to edit the instance of a game where I am running into problems. My view is calling the form = GameForm(request.POST, instance=game) where game = Game.objects.get(pk=id). The form is pre-populated with the correct data, but when it's submitted, whether there are updates or not, the form is not validating. It sees it as a POST, but cannot get inside the if form.is_valid() conditional. And this is ever since I added the user field. I am using the default Django User model, and the field name is called "owner." It is set up as a ManyToManyField(User, blank=True) as users can own many games, and games can be owned by many users. Django forms the Many-To-Many "through" table, but I don't want the user to be able to change who owns what. I have it as a hidden field in my forms.py so a user can't change it. Model class Game(models.Model): game_title = models.CharField(max_length=100, verbose_name='Game Title', db_column='game', blank=False, null=False, unique=True) game_developer = models.CharField(max_length=100, verbose_name='Developer', db_column='developer', blank=True, null=True) game_release = models.DateField(max_length=50, verbose_name='Release Date', db_column='release_date', blank=False, null=True) … -
Django: GenericInline combined with ModelMultipleChoiceField
Hi I'm working on an adminform which allows either selecting an existing object or add a new one in an inline of a GenericRelation. See the following example Models: from stdimage class Photograph(DeepZoom): content_type = models.ForeignKey(ContentType, on_delete=models.CASCADE, null=True) object_id = models.PositiveIntegerField(default=0) profile = GenericForeignKey("content_type", "object_id") img = JPEGField(...) img_alt = models.Charfield(max_lenght=300) ... and class MyProfile(models.Model): name = models.CharField(max_length=200, default="") photos = GenericRelation(Photograph, null=True) To achieve "select existing" I wrote a custom form for the admin site class SomeForm(forms.ModelForm): choose_img = forms.ModelMultipleChoiceField(queryset=Photograph.objects.all(), required=False) def __init__(self, *args, **kwargs): super(SomeForm, self).__init__(*args, **kwargs) self.initial.update({"choose_img": self.instance.photos.all()}) def save(self, commit=True): chosen_imgs = set(self.cleaned_data.pop("choose_img")) instance = super(SomeForm, self).save(False) if commit: instance.save() current_imgs = set(instance.photos.all()) for img in chosen_imgs - current_imgs: instance.photos.add(img) for old_img in current_imgs - chosen_imgs: old_img.profile = None old_img.object_id = 0 old_img.save() return instance and add it with the PhotographInline (basically just an instance of GenericTabularInline) to the admin site class MyProfileAdmin(admin.ModelAdmin): form = SomeForm inlines = [ PhotographInline, ] So far so good. The new save method does what i expect when selecting Photograph's in the choose_img field. If I deselect a Photograph it still works but it redirects me to the admin change form displaying "Please correct the errors below." and marking the … -
How to fix compatibility between libraries
Im trying to perform some actions with jwt token. Im working with Django-rest-jwt, wich requires PyJWT and jwt. The problem is - I made fork to PyJWT cause I need some code working another way, and when I try to use Django-rest-jwt, it says that he can't find jwt, also pip3 gives me this, when I install my fork. pic , btw fork works correctly. So, how can I make Django-rest-jwt use my fork instead of PyJWT? I tried to fork it and change requirements, but didn't find PyJWT at all -
Gunicorn "error: unrecognized arguments: --log-file--"
I am trying to deploy my django application to heroku, but every single time I try, I see this error in my logs: gunicorn: error: unrecognized arguments: --log-file-- But my Procfile looks like this (I haven't forgotten to save it or anything) web: gunicorn post_project.wsgi --log-file - I can't really put my eye on anything that may be a cause of that error except these few lines at Heroku This app is using free dynos web gunicorn post_project.wsgi --log-file-- -
Problem with sending a POST request with axios in reactjs
Hi every body I am using reactjs and redux for frontend and I want to create new articles so I created CreateArticle.js as below import { useDispatch, useSelector } from 'react-redux' import { unwrapResult } from '@reduxjs/toolkit' import { addNewArticle } from './managementSlice' import { selectAccessToken, selectUser } from '../authentications/authenticationsSlice' import './CreateArticle.scss'; import { selectAllSubjects, fetchSubjects } from '../subjects/subjectsSlice'; import { selectAllTags, fetchTags } from '../tags/tagsSlice'; export const CreateArticle = () => { const [title, setTitle] = useState('') const [body, setBody] = useState('') const [slug, setSlug] = useState('') const [subjectId, setSubjectId] = useState(0) const [tags, setTags] = useState([]) const [addRequestStatus, setAddRequestStatus] = useState('idle') const user = useSelector(selectUser); const subjects = useSelector(selectAllSubjects) const allTags = useSelector(selectAllTags) const subjectStatus = useSelector((state) => state.subjects.status) const tagStatus = useSelector( (state) => state.tags.status) const dispatch = useDispatch() // const onSubjectChanged = (e) => setSubjectId(parseInt(e.target.value)) const onSubjectChanged = (e) => setSubjectId(e.target.value) const onTitleChanged = (e) => { setTitle(e.target.value) setSlug(e.target.value.toLowerCase())} const onBodyChanged = (e) => { setBody(e.target.value) console.log(e.target.value); } const onTagsChanged = (e) => { const selectedOptions = [...e.target.selectedOptions].map(o => parseInt(o.value)) setTags(selectedOptions) } const onSaveArticleClicked = async () => { let article = { Subject:subjectId, Title: title, Body:body, Tags:tags, Slug:slug } if (user){ try { setAddRequestStatus('pending') console.log('addRequestStatus:', … -
How to import all the classes in a module as a list?
In django, to display models on admin page, we have to first register all the models in admin.py. If there are so many models to register, we have to write admin.site.register(model_name) so many times. Its kind of a boring task. So I was wondering if there is any way in python to import all the classes in module as list. for instance, admin.py, from django.contrib import admin from .models import * as list_name #like this is there any way we can get all things inside * in list ? for i in list_name: admin.site.register(i) #so that this way I have to write only one line of code to register all classes. Is there any way to save writing lot of lines here ? -
django - how to save form fields from multiple lines of input?
I want to add mutliple values from user input to a form field into one variable. These values are ManyToMany field in my model. User selects them from a list and adds into that field. I send {{form.product}} to template, I can gather one value from here but i want to add more lines (which i can with JS) and save those lines into same (forms.product) field. My model contains product,product_group. I want to add products with product group as well, which will be different in every line. E.g.: prod1,1 / prod1,2 / prod,3 etc. Second value will be that line index that i want to add. My template html looks like this: {{ form.product}} <ul>Products:</ul> <div id="productcon"> </div> <div id="add" onclick="add()">add interest</div> And my javascript is like this: function add(){ let index = $('#productcon')[0].childElementCount +1; $('#productcon').append(`<input type="text" name="product${index}" value="??"/> `); } This add new lines but I can't save them except my first value. This JS part confused me. Or i focused on JS and missing something else. I hope I could explain my problem. Thank you. -
get highest number of order on a single date
My model is like class Order(models.Model): user=models.ForeginKey(User) completion_date_time = models.DateTimeField(blank=True, null=True) I want to get date on which max number of order placed with count of order -
Return multiple user in Django
My tests are failing saying get() returned more than one permissions -- it returned 2! However I want to return more than one permission, but I could not get the list. I'm using ajax to display those list on modal. My problem is I want to display all usertype name base on idd below. If any expert could help please share some solution thanks in advance. returned more than one permissions -- it returned 2! views.py def ajax(request): idd = request.GET.get('id') app_permission_tbl_usid = permissions.objects.get(user_id=idd) app_permission_tbl_permid = app_permission_tbl_usid.permission_id datas = usertypes.objects.get(id=app_permission_tbl_permid) list_of_permission = datas.usertype return JsonResponse({"result": list_of_permission}) -
Render a different form inside view Django
I'd like to use the same page, but render a different form according to which button I press. Is there a way of doing it? Here it goes my HTML and VIEW code _adm.html <div class="btn-group btn-group-lg" role="group"> <div class="col-md-12 text-center"> <form action="" method="POST"> <a role="button" type="submit" href="{% url 'administrativo' user.username %}" name="senha" class="btn btn-green text-white">Alterar Senha</a> <a role="button" type="submit" href="{% url 'administrativo' user.username %}" name="nome" class="btn btn-green text-white">Alterar Nome</a> </form> </div> adm.html {% extends 'base.html' %} {% load static %} {% block content %} {% include 'partials/_header.html' %} {% include 'partials/_adm.html' %} <div class="container"> {{ form }} </div> {% include 'partials/_footer.html' %} {% endblock %} view.py def alterar_senha(request, user_username): usuario = User.objects.get(username=user_username) form = AlterarSenha(instance=usuario) context = { 'form': form } return render(request, 'usuarios/administrativo.html', context) def editar_usuario(request, user_username): usuario = User.objects.get(username=user_username) form = EditarUsuario(instance=usuario) context = { 'form': form } return render(request, 'usuarios/administrativo.html', context) def administrativo(request, user_username): if request.method == 'POST' and 'senha' in request.POST: return redirect('alterar_senha', user_username) elif request.method == 'POST' and 'nome' in request.POST: return redirect('editar_usuario', user_username) return render(request, 'usuarios/administrativo.html') I have no idea if it is possible to do such a thing. Otherwise, I'll have to render 3 different pages. -
How to change the default port of frontend and backend pages?
I have a front-end developed using react. I need this front-end page to run on port 1111. And also I have a back-end diango api which runs on a port 2222. the front-end receive data from the back-end. How can I configure the ports for both page. Note that for the back-end, i just added the following two lines from django.core.management.commands.runserver import Command as runserver runserver.default_port = "2222" to manage.py and now it is ok. But the frontend I get always localhost:3000 -
get item and its last reverse relationship item
How could i get joined results of item and only last reversed item using one query. for example class Blog: name = models.CharField() class Entry: blog = models.ForeignKey(Blog) how to get a list of distinct blog items with only last entry associated with it, thanks! -
is there way to create box for info or to be title in admin panel in django
is there way to create box for info or to be title in admin panel in django like photo for example when i add new record i want put box for info my admin.py : from django.contrib import admin from blog_app.models import Android # Register your models here. admin.site.register(Android) -
join() argument must be str, bytes, or os.PathLike object, not 'function' problem
I am currently working on my first website, but I am experiencing some issues such as this one: join() argument must be str, bytes, or os.PathLike object, not 'function'. I think it has something to do with views.py but I don't really know how to solve it. Here's the code: def translator_view(request): def translator(phrase): translation = "" for letter in phrase: if letter.lower() in "a": if letter.isupper: translation = translation + "U" else: translation = translation + "u" elif letter.lower() in "t": if letter.isupper: translation = translation + "A" else: translation = translation + "a" elif letter.lower() in "c": if letter.isupper: translation = translation + "G" else: translation = translation + "g" elif letter.lower() in "g": if letter.isupper: translation = translation + "C" else: translation = translation + "c" return translation return render(request, translator, 'main/translator.html') def translated_view(self, request): text = request.GET.get('text') print('text:', text) translator = self.translator dt = translator.detect(text) tr = self.translated.text context = { 'translated': tr } return render(request, context, 'main/translated.html') def home_view(request): return render(request,'main/home.html') -
how to add dynamic dropdown menu option in django
admin.py from django.contrib import admin from .models import ReportModel # Register your models here. @admin.register(ReportModel) class ReportAdmin(admin.ModelAdmin): list_display=('url', 'width', 'height', 'name') forms.py from django import forms from .models import ReportModel from django.db import models class ReportForm(forms.Form): width = forms.CharField(widget=forms.NumberInput(attrs={'class':' form-control'})) height = forms.CharField(widget=forms.NumberInput(attrs={'class':' form-control'})) url = forms.URLField(max_length=300) name = forms.CharField(max_length = 50) class Meta: model = ReportModel fields = [ "url","width","height","name"] urls.py from django.contrib import admin from django.urls import path, re_path from django.conf.urls import url from .import views app_name = 'report' urlpatterns = [ url(r'^reporttest/$', views.reporttest, name='reporttest'), url(r'^add/$', views.add, name='add'), ] view.py from .forms import ReportForm from .models import ReportModel from django.shortcuts import render, redirect def reporttest(request): form = ReportForm(request.POST) return render(request, 'report/add_report.html',{'form':form}) def add(request): report_items={} form = ReportForm(request.POST) url =request.POST['url'] width=request.POST['width'] height=request.POST['height'] name=request.POST['name'] if form.is_valid(): if ReportModel.objects.filter(url=url).exists(): return render(request, 'report/add_report.html', { 'form': form, 'error_message': 'URL already exists.' }) elif ReportModel.objects.filter(name=name).exists(): return render(request, 'report/add_report.html', { 'form':form, 'error_message': 'Report name already exists.' } ) else: report_models=ReportModel(url=url,width=width,height=height,name=name) report_models.save() print(report_models.name) report_items={'url':request.POST['url'], 'width':request.POST['width'], 'height':request.POST['height'], 'name':request.POST['name']} return render(request, 'report/report_one.html', report_items) else: return render(request, 'report/add_report.html',{'form':form}) report_one.html {% block content %} <body> <iframe src = "{{url}}" width= "{{width}}" height= "{{height}}" frameborder="0" allowfullscreen allowtransparency ></iframe> </body> {% endblock %} add_report.html {% extends "student/base2.html" %} {% block content %} … -
Can I link my customer model to my user model?
I'm working an a django e-commerce website where a user has to be a customer. But when I create a new user, it assigns it to the the superuser not the new user and get this error: Exception Value: User has no customer. But i can also go to my admin panel and re-assign the customer to the user. django admin panel How can I fix this please? My customer model class Customer(models.Model): user = models.OneToOneField(settings.AUTH_USER_MODEL, on_delete=models.CASCADE) name = models.CharField(max_length=255, null=True) email = models.CharField(max_length=255, null=True) def __str__(self): return self.name members/views.py to create new user and customer from django.shortcuts import render from django.views import generic from django.contrib.auth.forms import UserCreationForm from django.urls import reverse_lazy from shop.models import Customer from django.views.generic import CreateView from .forms import CreateCustomerForm # Create your views here. class UserRegisterView(generic.CreateView): form_class = UserCreationForm template_name = 'registration/signup.html' success_url = reverse_lazy('customer') class CreateCustomerView(CreateView): form_class = CreateCustomerForm template_name = 'registration/customerProfile.html' success_url = reverse_lazy('login') def form_valid(self, form): form.instance.user = self.request.user return super().form_valid(form) My shop view def shop(request): data = cartData(request) cartItems = data['cartItems'] products = Product.objects.all() context = {'products': products, 'cartItems': cartItems} return render(request, 'shop/shop.html', context)
