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我正在尝试使用基于函数的视图创建详细视图。此视图必须仅显示已发布的帖子和非草稿帖子。

def singlePost(request, slug_post, slug_category):
    post_category = get_object_or_404(BlogCategory, slug_category=slug_category)
    if post_filter == BlogPost.objects.filter(draft=True):
        raise PermissionDenied
    if post_filter == BlogPost.objects.filter(publishing_date__gt=datetime.datetime.now()):
        raise PermissionDenied
    else:
        post_filter == BlogPost.objects.all()
    post_details = get_object_or_404(post_filter, slug_post=slug_post)
    category_post_details = BlogPost.objects.filter(post_category=post_category)
    context = {
        "post_category": post_category,
        "post_details": post_details,
        "category_post_details": category_post_details,
        }
    template = 'blog/reading/single_post.html'
    return render(request, template, context)

但是当我使用它时，我看到了这个错误消息：

名称“post_filter”未定义

我该如何解决？

注意：视图以这种方式工作正常

def singlePost(request, slug_post, slug_category):
    post_category = get_object_or_404(BlogCategory, slug_category=slug_category)
    post_details = get_object_or_404(BlogPost, slug_post=slug_post)
    category_post_details = BlogPost.objects.filter(post_category=post_category)
    context = {
        "post_category": post_category,
        "post_details": post_details,
        "category_post_details": category_post_details,
        }
    template = 'blog/reading/single_post.html'
    return render(request, template, context)
4

1 回答 1

1

根据提供的信息，我建议使用以下方法。

使用单个过滤器获取博客文章的条件，如果不存在则引发错误。

post_filter = BlogPost.objects.filter(draft=False,
                                      publishing_date__lt=datetime.datetime.now()):
if not post_filter.exists():
    raise PermissionDenied
else:
    post_details = get_object_or_404(post_filter, slug_post=slug_post)
于 2019-07-15T16:03:01.067 回答