Open listView with foreignkey from another listview with Django - django

Question

I am working on a product overview page in Django.
For this I have three models: category, brand, product.
I have created a View with ListView of the category. I loop through them to display them. I then want to open another overview of all brands within that category.
How do I do this?
Here are my models:
class Category(models.Model):
category_name = models.CharField(max_length=200)
sub_category = models.CharField(max_length=200,blank=True,null=True)
category_picture = models.ImageField(upload_to='category/', null=True, blank=True)
def __str__(self):
if self.sub_category is None:
return self.category_name
else:
return f" {self.category_name} {self.sub_category}"
class Meta:
ordering = ['category_name']
class Brand(models.Model):
category = models.ForeignKey('Category', on_delete=models.SET_NULL,null=True,blank=True)
brand_name = models.CharField(max_length=200)
brand_owner = models.CharField(max_length=200)
brand_story = models.TextField()
brand_country = models.CharField(max_length=200)
def __str__(self):
return f"{self.brand_name}"
class Bottle(models.Model):
category_name = models.ForeignKey('Category', on_delete=models.SET_NULL,null=True,blank=True)
brand = models.ForeignKey('Brand', on_delete=models.CASCADE)
bottle_name = models.CharField(max_length=255)
bottle_info = models.TextField()
bottle_tasting_notes = models.TextField()
bottle_barcode = models.IntegerField()
bottle_image = models.ImageField(upload_to='bottles/',null=True)
def __str__(self):
return f"{self.brand.brand_name} {self.bottle_name}"
How do I open a listview of all brands within a certain category from a link of the category listview?

Answer 1

First thing is to create another listview that displays all brands within a specific category. We do this by creating a new get_queryset() method.
views.py
class BrandListView(ListView):
model = Brand
def get_queryset(self):
return Brand.objects.filter(category__category_name=self.kwargs['category'])
Next add a url to your URLS.py so it can be accessed. We're using category_name as part of the url so it's human readable
from .views import BrandListView
urlpatterns = [
path('brands/<str:category>/', PublisherBookList.as_view()), name= "brand_list"
]
Then, as you loop through your categories in your template, create a link to the url
{% for category in categories %}
{{category.category_name}} : See brands in category
{% endfor %}
This will work as long as your categories have fairly simple names. If not, you might want to use the ID in the URL instead, or add a slug field to the model and use that.