Question

了解Django，并使用Django书来跟踪探戈，但最后一个问题在添加子类别之后得到了我，该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)

这一切都很有效，直到我添加具有相同名称的子类别，例如子类别＆＃34;其他＆＃34;适用于多个类别。

我理解为什么，当我到达＆＃34; def subategory＆＃34;我的slug将返回多个子类别，因此我需要以某种方式将这些限制为相关类别，例如

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

或者什么;）

不确定最佳路线是什么，以及如何过滤这些

Answer 1

鉴于您可能为两个不同的SubCategory对象设置了两个具有相同名称的Category个对象，如您所建议的那样，您可以添加Category作为附加过滤器。

按SubCategory.slug和Category.slug

进行过滤

为了达到这个目的，我看到你有一个视图，它为SubCategory和Category设置了slu，你定义的像这样subcategory(request, subcategory_name_slug, category_name_slug)。这些就足以过滤：

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

您在上面看到我使用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（...）。

每个类别

实施SubCategory.name的 unicity

要使用get()安全地执行此操作，需要保证对于任何给定的类别，不会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
        )

django子类slug过滤器

1 个答案:

按SubCategory.slug和Category.slug

每个类别