CODEWITHSUNDEEP
pythondjangourldjango-views
KamWebDev
KamWebDev

Reputation: 377

Django 2.0 url parameters in get_queryset

I would like to filter subcategories based on the category id from the URL

For a constant value, it works without a problem

return Subcategory.objects.filter(category = 1)

views.py

class SubcategoriesListView(ListView):
    model = Subcategory
    template_name = 'app/categories/index.html'
    def get_queryset(self):
        return Subcategory.objects.filter(category = category_id)

urls.py

path('categories/<int:category_id>/', app.views.SubcategoriesListView.as_view(), name='subcategories'),

models.py

class Subcategory(models.Model):
   title = models.CharField(max_length=30)
   category = models.ForeignKey(Category, on_delete=models.CASCADE)

Traceback

NameError at /categories/1/
name 'category_id' is not defined

views.py in get_queryset
return Subcategory.objects.filter(category = category_id)

Upvotes: 18

Views: 7331

Answers (1)

willeM_ Van Onsem
willeM_ Van Onsem

Reputation: 476813

You can obtain the URI positional and named parameters in a class-based view with self.args (a tuple) and self.kwargs (a dictionary) respectively.

Here you defined the category_id as a named parameter, so you can obtain its corresponding value with self.kwargs['category_id']:

class SubcategoriesListView(ListView):
    model = Subcategory
    template_name = 'app/categories/index.html'
    def get_queryset(self):
        return Subcategory.objects.filter(category_id=self.kwargs['category_id'])

Since the id is an integer, you thus filter on category_id, not on category.

This is detailed in the "dynamic filtering" section of Django's documentation on Built-in class-based generic views.

Upvotes: 25

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