獲取所有的父類的Django - 跟隨外鍵UP

Question

我有以下型號：

class Category(models.Model): 
    name = models.CharField(max_length=255) 
    parent = models.ForeignKey("self", blank=True, null=True) 

    class Meta: 
     verbose_name = _("category") 
     verbose_name_plural = _("categories") 

    def __unicode__(self): 
     return self.name 


class Item(models.Model): 
    name = models.CharField(max_length=100, verbose_name=_("name")) 
    keywords = models.CharField(max_length=255, verbose_name=_("keywords")) 
    category = models.ForeignKey(Category) 

    class Meta: 
     abstract = True 
     verbose_name = _('item') 
     verbose_name_plural = _('items') 


class Product(Item): 
    price = models.DecimalField(decimal_places=2, max_digits=8, verbose_name=_("price")) 
    brand = models.ForeignKey(Brand, verbose_name=_("brand")) 
    article_number = models.CharField(max_length=255, verbose_name=_("article_number")) 

    def __unicode__(self): 
     return self.name 

    class Meta: 
     verbose_name = _('product') 
     verbose_name_plural = _('products') 

比方說，我在數據庫以下幾類：

ID  NAME  PARENT_ID 
1  Products  null 
2  Phones   1 
3  iPhones   2 

我可以得到頂級類通過執行以下操作：

#This is a product with the category "iPhones" 
product.category.parent.parent 

但這並不好，因爲產品可以是x個類別的深度。

如何獲取數組或其他類中的所有相關類？

Wanted output = [iPhones, Phones, Products] 

Answer 1

寫模型屬性方法的項目類：

class Item(models.Model): 
    @property 
    def all_categories(self): 
     categories = [] 
     current_category = self.category 
     while current_category is not None: 
      categories.append(current_category) 
      current_category = current_category.parent 
     return categories 
     #optional return reversed list: 
     #return list(reversed(categories)) 

現在你可以用

product.all_categories