Question

我有两个带有slug字段的模型：

class Book(models.Model):
    name = models.CharField(max_length=200)
    slug = models.SlugField()

class Author(models.Model):
    name = models.CharField(max_length=200)
    slug = models.SlugField()

我想将它们映射到第一级路径：

(r'^(?P<slug>[a-zA-Z0-9_-]+)/$', 'book_detail'),
(r'^(?P<slug>[a-zA-Z0-9_-]+)/$', 'author_detail'),

如果不使用相同的功能并根据slug返回书籍或作者，最好的方法是什么。

Answer 1

最好的方法是在视图中将其拆分：

r'^(?P<model>[a-zA-Z0-9_-]+)/(?P<slug>[a-zA-Z0-9_-]+)/$', 'some_detail')

并查看：

def some_detail(request, model, slug):
    try:
        model = {'book':Book, 'author':Author}[model]
    except KeyError:
        raise Http404

    item = get_object_or_404(model, slug=slug)
    do_something_with(item)
    ...

编辑：哦，那样平坦......那将是：

(r'^(?P<slug>[a-zA-Z0-9_-]+)/$', 'universal_detail'),

def universal_detail(request, slug):
    try:
        book = Book.objects.get(slug=slug)
        return book_detail(request, book)
    except Book.DoesNotExist:
        pass

    try:
        author = Author.objects.get(slug=slug)
        return author_details(request, author)
    except Author.DoesNotExist:
        raise Http404

 def book_detail(request, book):
    # note that book is a book instance here
    pass

将相同的第一级URL路径映射到不同的模型

1 个答案: