У меня в статусе модели с Dropdown:
models.py
class Item(models.Model):
DRAFT = "D"
PUBLISHED = "P"
CLOSED = "C"
STATUS = (
(DRAFT, _("Draft")),
(PUBLISHED, _("Published")),
(CLOSED, _("Closed")),
)
user = models.ForeignKey(
settings.AUTH_USER_MODEL, null=True, related_name="articleauthor",
on_delete=models.SET_NULL)
...
status = models.CharField(max_length=1, choices=STATUS, default=DRAFT)
...
forms.py
class ItemForm(forms.ModelForm):
#status = forms.CharField(widget=forms.HiddenInput())
edited = forms.BooleanField(widget=forms.HiddenInput(), required=False, initial=False)
content = MarkdownxFormField()
class Meta:
model = Item
fields = ["title", "content", "image", "price", "currency", "status", "edited"]
Но теперь я хочу, чтобы при создании новых предметов удалялась закрытая запись.
views.py
class CreateItemView(LoginRequiredMixin, PermissionRequiredMixin, CreateView):
"""Basic CreateView implementation to create new articles."""
permission_required = 'is_staff'
model = Item
message = _("Your article has been created.")
form_class = ItemForm
template_name = 'item_create.html'
def form_valid(self, form):
form.instance.user = self.request.user
return super().form_valid(form)
def get_success_url(self):
messages.success(self.request, self.message)
return reverse('offers:list')
'''
def get_queryset(self):
form = self.form = self.form_class(self.request.GET)
if self.request.user.is_authenticated():
form.fields['status'].choices.pop("C")
form.fields['status'].widget.choices.pop("C")
'''
как бы вы специально скрывали "закрытое" для одного представления?