python - django 子类别 slug 过滤器

Question

了解 Django 并遵循 Django 书的探戈，但最后一期在添加子类别后让我困惑，这些子类别未包含在该教程中。

我有以下内容:

models.py
class Category(models.Model):
"""Category"""
name = models.CharField(max_length=50)
slug = models.SlugField()


def save(self, *args, **kwargs):

                    #self.slug = slugify(self.name)
            self.slug = slugify(self.name)
            super(Category, self).save(*args, **kwargs)
def __unicode__(self):
    return self.name


class SubCategory(models.Model):
"""Sub Category"""
category = models.ForeignKey(Category)
name = models.CharField(max_length=50)
slug = models.SlugField()

def save(self, *args, **kwargs):

            self.slug = slugify(self.name)
            super(SubCategory, self).save(*args, **kwargs)

def __unicode__(self):
    return self.name

和

urls.py
(r'^links/$', 'rango.views.links'),
(r'^links/(?P<category_name_slug>[\w\-]+)/$', 'rango.views.category'),  
(r'^links/(?P<category_name_slug>[\w\-]+)/(?P<subcategory_name_slug>[\w\-]+)/$', 'rango.views.subcategory'),  

和

views.py
@require_GET
def links(request):
"""Linkdirectory Page"""
category_list = Category.objects.order_by('name')
context_dict = {'categories': category_list}
return render(request, 'links.html', context_dict)

@require_GET
def category(request, category_name_slug):
"""Category Page"""
category = Category.objects.get(slug=category_name_slug)
subcategory_list = SubCategory.objects.filter(category=category)
context_dict = {'subcategories': subcategory_list}
return render(request, 'category.html', context_dict)

@require_GET
def subcategory(request, subcategory_name_slug, category_name_slug):
"""SubCategory Page"""
context_dict = {}
try:
    subcategory = SubCategory.objects.get(slug=subcategory_name_slug)
    context_dict['subcategory_name'] = subcategory.name
    websites = Website.objects.filter(sub_categories=subcategory)
    context_dict['websites'] = websites
    context_dict['subcategory'] = subcategory
except SubCategory.DoesNotExist:
return render(request, 'subcategory.html', context_dict)

这一切都运行良好，直到我添加具有相同名称的子类别，例如多个类别的子类别“其他”。

我明白为什么，当我到达“def subcategory”时，我的 slug 将返回多个子类别，因此我需要以某种方式将这些子类别限制为相关类别，例如

"SELECT 
subcategory = SubCategory.objects.get(slug=subcategory_name_slug)
WHERE 
subcategory = SubCategory.objects.filter(category=subcategory)
CLAUSE" 

或者什么;)

不确定采取什么最佳途径以及如何过滤这些

Answer 1

鉴于您可能有两个不同的 SubCategory 对象，它们对于两个不同的 Category 对象具有相同的名称，正如您所建议的，您可以添加 Category > 作为附加过滤器。

按 SubCategory.slug 和 Category.slug 过滤

为了实现这一目标，我看到您有一个 View ，它采用 SubCategory 和 Category 的别名，您像这样定义subcategory(request, subcategory_name_slug, Category_name_slug )。这些足以过滤:

subcategory = SubCategory.objects.get(
    slug=subcategory_name_slug,
    category__slug=category_name_slug
)
           ^
           |__ # This "double" underscore category__slug is a way to filter
               # a related object (SubCategory.category)
               # So effectively it's like filtering for SubCategory objects where
               # SubCategory.category.slug is category_name_slug

如上所示，我使用 SubCateogry.objects.get(...) 来获取单个对象，而不是可以返回许多对象的 `SubCategory.objects.filter(...) .

强制每个类别的 SubCategory.name 的唯一性

要使用 get() 安全地执行此操作，需要保证对于任何给定类别，不超过一个具有相同名称的子类别

您可以使用 unique_together 强制执行此条件

class SubCategory(models.Model):
    class Meta:
        unique_together = (
            ('category', 'name'),          # since slug is based on name,
                                           # we are sure slug will be unique too
        )