Django

Question

我正在尝试包含一个模板并在两个不同的视图中使用它，其中第一个获得一个 url 参数，第二个获得相同的参数加上另一个。在我包含的模板中有一个带有 {% url %} 标签的迭代，我需要传递这两个参数，因为第二个视图需要它们，但是这样做会在尝试呈现我的第一个视图时导致 NoReverseMatch，可能是因为它只接受一个参数。有没有办法指定第二个参数是可选的？

这是我的代码：

urls.py

...

    url(r'^portfolio/(?P<category_slug>[-\w]+)/$', views.category, name='category'),
    url(r'^portfolio/(?P<category_slug>[-\w]+)/(?P<album_slug>[-\w]+)/$', views.album, name='album'),

...

models.py

...

class Album(models.Model):
    cover = models.ImageField()
    title = models.CharField(max_length=200, unique=True)
    description = models.CharField(max_length=200, blank=True)
    posts = models.ManyToManyField(Post, blank=True)
    slug = models.SlugField(max_length=200)


class Category(models.Model):
    cover = models.ImageField()
    title = models.CharField(max_length=200, unique=True)
    albums = models.ManyToManyField(Album, blank=True)
    slug = models.SlugField(max_length=200)


@receiver(pre_save, sender=Album)
@receiver(pre_save, sender=Category)
def save_slug(sender, instance, *args, **kwargs):
    instance.slug = slugify(instance.title)

...

views.py

    ...

    def index(request):
        main_categories = Category.objects.filter(...)
        return render(request, 'index.html', {'main_categories': main_categories})

    def category(request, category_slug):
        try:
            category_selected = Category.objects.get(slug=category_slug)
        except Category.DoesNotExist:
            category_selected = None
        albums = category_selected.albums.all()
        return render(request, 'category.html', {
            'category_selected': category_selected,
            'albums': albums
        })


    def album(request, category_slug, album_slug):
        try:
            category_selected = Category.objects.get(slug=category_slug)
            album_selected = Album.objects.get(slug=album_slug)
        except Category.DoesNotExist:
            category_selected = None
        except Album.DoesNotExist:
            album_selected = None
        posts = album_selected.posts.all()
        return render(request, 'album.html', {
            'category_selected': category_selected,
            'album_selected': album_selected,
            'posts': posts
        })

    ...

includedtemplate.html

...

{% for obj in objects %}

    ...    

    <a href="{% url view category_slug=obj.slug album_slug=obj.slug %}">

    ...

{% endfor %}

...

index.html

...

{% include 'includedtemplate.html' with objects=main_categories view='category' %}

...

编辑： 我设法通过将我的 urls 与每个不同的 slug 分开来解决这个问题。这更简单，更适合我的情况，考虑到我有一个 M2M 用于类别和专辑，这可能会导致单个专辑有很多 url。

Answer 1

您可以合并视图并为 album_slug 设置 None。

def combined_view(request, category_slug, album_slug=None):

    category_selected = Category.objects.filter(slug=category_slug).first()
    album_selected = None
    posts = None
    template = 'category.html'

    if album_slug:
        album_selected = Album.objects.filter(slug=album_slug).first()
        posts = album_selected.posts.all()
        template = 'album.html'

    return render(request, template, {
        'category_selected': category_selected,
        'album_selected': album_selected,
        'posts': posts
    })

urls 的顺序也很重要 - 第一个 url 应该有一个参数，第二个有两个参数：

url(r'^portfolio/(?P<category_slug>[-\w]+)/$', views.combined_view, name='cview'),
url(r'^portfolio/(?P<category_slug>[-\w]+)/(?P<album_slug>[-\w]+)/$', views.combined_view, name='cview'),

P.S。在视图中，您可以使用 filter().first() 而不是 try..except。写起来更快。

Django - 包含模板的多个 url 参数

Django - multiple url params with included template

python

django-templates

django-urls

urls.py

models.py

views.py

includedtemplate.html

index.html