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我有一个使用模板如下列表视图(用于管理界面):如何将Django ListView转换为管理网站上的工作?
class UserImageListPendingView(ListView):
model = UserImage
queryset = UserImage.objects.filter(status=ImageBase.PENDING)
template_name = 'userimage_list_pending.html'
context_object_name = 'userimage_list'
paginate_by = 5
@method_decorator(staff_member_required)
def dispatch(self, *args, **kwargs):
return super(UserImageListPendingView, self).dispatch(*args, **kwargs)
虽然这个工程有问题,把网址中的urls.py:
urlpatterns = [
url(r'^admin/app/pendinguserimages/?$', login_required(
UserImageListPendingView.as_view()),
name='pendinguserimages'),
...
]
。 ..这会阻止重定向正常工作。
我曾尝试通过admin.py定义的网址:
def get_admin_urls(urls):
def get_urls():
return patterns('',
url(r'^app/pendinguserimages/?$',
UserImageListPendingView.as_view(), name='pendinguserimages'),
url(r'^app/checkuserimage/(?P<pk>[0-9]+)/?$',
userimage_check, name='checkuserimage'),
...
) + urls
return get_urls
admin_urls = get_admin_urls(admin.site.get_urls())
admin.site.get_urls = admin_urls
...但有扭转checkuserimage URL时出错。
我该如何转换此视图以适应管理网站,但仍然使用该模板?