Python 如何在与类别相关的文章上设置正确的绝对url

我用它来制作一个自定义类别模型

我有一个与类别相关的文章模型，但是如何在文章模型上设置get_absolute_url，比如/category path/article slug

/类别路径/可以是这样的路径：/top-level-category/sub-level-category/sub-sub-level-category/

这是我的模特

class Category(CategoryBase):

    order = models.IntegerField(default=0)
    description = models.TextField(blank=True, null=True)

    class MPTTMeta:
        order_insertion_by = ('order', 'name')

    def get_absolute_url(self):
        """Return a path"""
        prefix = reverse('categories_tree_list')
        ancestors = list(self.get_ancestors()) + [self, ]
        return prefix + '/'.join([force_unicode(i.slug) for i in ancestors]) + '/'


class Article(models.Model):
    title = models.CharField(max_length=255)
    slug = models.SlugField(max_length=255)

    category = models.ForeignKey(Category)

我的URL.py

from django.conf.urls import patterns, url
from apps.news.views import CategoryList


urlpatterns = patterns('',
    #Startpage for app
    url(r'^$', CategoryList.as_view(), name="categories_tree_list")
)


urlpatterns += patterns('apps.news.views',
    url(r'^(?P<path>.+)/$', 'category_detail', name='categories_category'),
)

我不知道类别是否有属性slug，如果没有，您可以自己添加它

models.py

class Category(CategoryBase):

    order = models.IntegerField(default=0)
    description = models.TextField(blank=True, null=True)

    class MPTTMeta:
        order_insertion_by = ('order', 'name')

    def get_absolute_url(self):
        """Return a path"""
        prefix = reverse('categories_tree_list')
        ancestors = list(self.get_ancestors()) + [self, ]
        return prefix + '/'.join([force_unicode(i.slug) for i in ancestors]) + '/'


class Article(models.Model):
    title = models.CharField(max_length=255)
    slug = models.SlugField(max_length=255)

    category = models.ForeignKey(Category)

class Article(models.Model):
    ...

    category = models.ForeignKey(Category)

    def get_absolute_url(self):
        arguments = (self.category.slug, self.slug)
        return reverse('article-detail',arguments)

views.py

def article_detail(request,category_slug,article_slug):
    article = Article.objects.get(category__slug=category_slug, 
                                  slug=article_slug)
    return render(request, 'article_detail_template.html', 
                  {'article':article})

url.py

urlpatterns += patterns('',
    url(r'^(?P<category_slug>.+)/(?P<article_slug>.+)$', 
            'app.views.article_detail', name='article-detail'),
)

在模板中，使用以下链接到特殊文章

---

是的，如果这篇文章属于顶级类别，那就太好了。当文章处于子级别类别（如/top category slug/sub category slug/article slug/I有一个建议，你的文章的slug不以/）结尾时，这种方法不起作用，那么很容易判断url是类别还是文章，很容易实现你的目标