为什么此代码不起作用,如何获取url的特定部分?

时间:2019-05-31 11:58:58

标签: python django

我有以下网址http://localhost:8000/enthusiastically-synergize-optimal-results-after-proactive-total-linkage/edit/,并希望在此函数中从该网址访问该条,以便我可以使用它。在基于类的视图中,我可以使用self.kwargs.get('smthng')并抓住它,但不知道我该怎么做:

def permission_of_user_for_posts(request):
post = get_object_or_404(Blog, slug= request.GET.get('blog_slug'))
if request.user.username == post.author:
    return True
return False


@user_passes_test(permission_of_user_for_posts)
def blog_update(request, blog_slug):
blog = get_object_or_404(Blog, slug=blog_slug)
if request.method == 'POST':
    form = BlogForm(request.POST, request.FILES, instance=blog)
    if form.is_valid():
        form.instance.author = request.user
        form.save()
        return redirect('/')

form = BlogForm(instance=blog)
context= {
    'form': form
}
return render(request, 'blogs/form.html', context)

还是在permission_of_user_for_posts函数中我做错了什么?请帮我谢谢你

编辑:这是网址

urlpatterns = [
    path('', blogs_view.home, name='home'),
    path('blogs/', blogs_view.blogs, name='blogs'),
    path('<slug:blog_slug>', blogs_view.blog_detail, name='blog_detail'),
    path('post/', blogs_view.blog_create, name='blog_create'),
    path('<slug:blog_slug>/edit/', blogs_view.blog_update, name='blog_update'),
    path('<slug:blog_slug>/delete/', blogs_view.blog_delete, name='blog_delete'),
]

1 个答案:

答案 0 :(得分:1)

执行此操作的方法是根本不使用装饰器或单独的函数。只需查看视图中的帖子即可:

def blog_update(request, blog_slug):
    blog = get_object_or_404(Blog, slug=blog_slug)
    if not request.user == blog.author:
        raise PermissionDenied 
    if request.method == 'POST':
        ...