Okay, your URL should point to your CategoryListView. That should have category_list in its context. And the template should in the category_list template.
From: django-users@googlegroups.com [mailto:django-users@googlegroups.com] On Behalf Of Jay Sent: Monday, August 27, 2018 10:28 AM To: Django users Subject: Re: Printing model fields organized by category Thanks Matthew, here is my views.py file: from django.shortcuts import render #generates HTML fiels using a template and data #from django.http import HttpResponse from .models import Book, Author, BookInstance, Genre, Model, ModelInstance, Category, Ownership, Manufacturer, Location #imports model classes to access data in views from django.views import generic # Create your views here. #def index(request): # return HttpResponse("Hey") def index(request): """View function for home page of site.""" #Generate counts of some of the main objects num_books = Book.objects.all().count() num_instances = BookInstance.objects.all().count() #Available books (status - 'a') num_instances_available = BookInstance.objects.filter(status__exact='a').count() #The 'all()' is implied by default. num_authors = Author.objects.count() #Num visits to this view, as counted in the session variable num_visits = request.session.get('num_visits', 0) request.session['num_visits'] = num_visits + 1 context = { 'num_books': num_books, 'num_instances': num_instances, 'num_instances_available': num_instances_available, 'num_authors': num_authors, 'num_visits': num_visits, } #Render the HTML template index.html with the data in the context variable return render(request, 'index.html', context=context) class BookListView(generic.ListView): model = Book class BookDetailView(generic.DetailView): model = Book class ModelListView(generic.ListView): model = Model class ModelDetailView(generic.DetailView): model = Model class CategoryListView(generic.ListView): model = Category class CategoryDetailView(generic.DetailView): model = Category from django.contrib.auth.mixins import LoginRequiredMixin class LoanedBooksByUserListView(LoginRequiredMixin,generic.ListView): """Generic class-based view listing books on loan to current user.""" model = BookInstance template_name ='catalog/bookinstance_list_borrowed_user.html' paginate_by = 10 def get_queryset(self): return BookInstance.objects.filter(borrower=self.request.user).filter(status__exact='o').order_by('due_back') On Monday, August 27, 2018 at 11:00:57 AM UTC-4, Matthew Pava wrote: We need to see your view code. I’m assuming now that you don’t have category_list in your context variable that you submitted to the template. From: django...@googlegroups.com<javascript:> [mailto:django...@googlegroups.com<javascript:>] On Behalf Of Jay Sent: Monday, August 27, 2018 9:20 AM To: Django users Subject: Re: Printing model fields organized by category Thanks for the suggestions, I will change the name from Model after I can get this working. So, I've changed the code using your help to: {% if category_list %} {% for category in category_list %} <li><strong>{{ category }}</strong></li> <ul> {% for model in category.model_set %} <li> <a href="{{ model.get_absolute_url }}"> {{ model.model_number }}..........{{ model.description }} </a> </li> {% endfor %} </ul> {% endfor %} {% else %} <p>There is no equipment in the database</p> {% endif %} {% endblock %} However, this just prints "There is no equipment in the database". Am I missing something? I will try to experiment a little more. On Monday, August 27, 2018 at 9:52:56 AM UTC-4, Matthew Pava wrote: Hi Jay, Firstly, I would avoid calling a model “Model.” Maybe “Product” would be better? It’s only because of Django’s models.Model class. That will likely cause confusion in the future for you. You’ll want to work with the Category model primarily and use a reverse lookup to get to the corresponding “Model” instead of working with a model_list. The reverse lookup of category to model is model_set by default. You can change the name if you want. {% for category in category_list %} <strong>{{ category }}</strong> <ul> {% for model in category.model_set %} <li><a href="{{ model.get_absolute_url }}">{{ model }}</a></li> {% endfor %} </ul> {% empty %} <p>There is no equipment in the database with a category.</p> {% endfor %} Check out https://docs.djangoproject.com/en/2.1/topics/db/examples/many_to_one/ From: django...@googlegroups.com<mailto:django...@googlegroups.com> [mailto:django...@googlegroups.com] On Behalf Of Jay Sent: Monday, August 27, 2018 8:02 AM To: Django users Subject: Printing model fields organized by category I have a Class called "Model" that has a field called "model_numbers". I have another class called "Category" that has a field called "category". In the Model Class, I have a field called category that is a Foreign Key to the field category in the Class category. [Screen Shot 2018-08-27 at 8.58.54 AM.png] Also, here is the html that prints the modeul_number and category fields: [Screen Shot 2018-08-27 at 9.01.10 AM.png] What the code is printing: CATEGORY 1: - Field 1 - Field 2 - Field 3 CATEGORY 2: - Field 1 - Field 2 - Field 3 What I want it to print is: CATEGORY 1: - Field 1 - Field 2 CATEGORY 2: - Field 3 In this way it prints out all the model numbers under their respective category name (instead of all model numbers under every category, which is not correct as each model number only has one category). I posted about this on stackoverflow, and received a response but I am still confused. <https://stackoverflow.com/questions/51882942/how-can-i-organize-a-list-of-fields-based-on-another-field> Thank you! -- You received this message because you are subscribed to the Google Groups "Django users" group. 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