[英]getting all products from parent category django rest
Toys
, Toys -> ChildToys
, Toys -> ChildToys -> Cars
我有相关产品的Toys -> ChildToys -> Cars
,我可以得到所有产品均通过Cars
,但我怎样才能通过过滤得到的产品Toys
类别。 到目前为止我所做的
class Category(models.Model):
name = models.CharField(max_length=100)
slug = models.SlugField(max_length=100, blank=True, unique=True)
icon = models.ImageField(upload_to='category_icons', null=True, blank=True)
parent = models.ForeignKey('self', on_delete=models.SET_NULL, related_name='children', null=True, blank=True)
class Product(models.Model):
category = models.ForeignKey(Category, blank=True, null=True, on_delete=models.CASCADE)
name = models.CharField(max_length=200)
class ProductList(generics.ListAPIView):
serializer_class = ProductSerializer
def get_queryset(self):
slug = self.request.query_params.get('slug', None)
parent_id = Category.objects.get(slug=slug)
cats = Category.objects.filter(parent_id=parent_id)
qs = Product.objects.filter(category_id__in=parent_id)
print(qs)
return qs
您可以使用以下方法过滤:
class ProductList(generics.ListAPIView):
serializer_class = ProductSerializer
def get_queryset(self):
slug = self.request.query_params.get('slug')
return Product.objects.filter(category__parent__slug=slug)
或在敏感匹配病例:
class ProductList(generics.ListAPIView):
serializer_class = ProductSerializer
def get_queryset(self):
slug = self.request.query_params.get('slug')
return Product.objects.filter(category__parent__slug__iexact=slug)
如果您需要下降整个子树,则需要在Category
模型上创建一个方法,以递归方式获取所有子类别。 还有另一篇文章向您展示了如何做到这一点。
所以假设你有这个方法Category.get_family_tree()
class Category(models.Model):
# fields
def get_family_tree(self):
# code from question linked above
# returns list of `Category` instances
那么在您看来,您可以像这样获取产品:
def get_queryset(self):
slug = self.request.query_params.get('slug')
categories = Category.objects.get(slug=slug).get_family_tree()
return Product.objects.filter(category__parent__in=categories)
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