Question

我正在使用django-categories制作自定义类别模型。

我有一个与该类别相关的文章模型，但我如何在文章模型上get_absolute_url像这样/category-path/article-slug

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

这是我的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)

我的urls.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'),
)

Answer 1

我不知道Category是否有slug属性，如果没有，你可以自己添加。

models.py

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})

urls.py

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

在模板中使用以下内容链接到特殊文章

<a href="{{ article_obj.get_absolute_url }}">{{ article_obj.title }}</a>

如何在与类别相关的文章上设置正确的绝对URL

1 个答案: