使用 FormView 保存在 Django 表单中

Question

I'm creating a form that will change the state of reserve book, I have this

    class LibraryReserveForm(CrispyFormMixin, forms.Form):

          def __init__(self, *args, **kwargs):
              self.manager = kwargs.pop('manager')
              super(LibraryReserveForm, self).__init__(*args, **kwargs)

          def save(self):
              self.instance.reserve_status = 'approved'

              self.instance.save()

              return self.manager

models.py

    class ReservedBooks(TimeStampedModel):

        BOOK_RESERVE_STATUS = Choices(
             ('for_approval', "For Approval"),
             ('approve', "Approved"),
             ('cancelled', "Cancelled"),
             ('rejected', "Rejected")
        )

        reserve_status = models.CharField(
               _('Status'),
               max_length=32,
               choices=BOOK_RESERVE_STATUS,
               default='for_approval'
        )

    ...

view

    class LibraryReserveView(
          ProfileTypeRequiredMixin,
          MultiplePermissionsRequiredMixin,
          FormView,
    ):

    model = ReservedBooks
    template_name = 'library/reserved_list.html'
    form_class = LibraryReserveForm

    def get_form_kwargs(self):
        kwargs = super(LibraryReserveView, self).get_form_kwargs()

        kwargs.update({
            'manager': self.request.user.manager,
        })

        return kwargs

urls

    url(
        r'^reserve/(?P<pk>\d+)/$',
        views.LibraryReserveView.as_view(),
        name='reserved'
    ),

everytime I submit the button I print something in the save() method of the forms but its not printing something therefore that method is not called. How do you called the save method ? Thanks

Answer 1

A FormView does not handle saving the object. It simply calls form_valid that will redirect to the success_url . But an UpdateView adds boilerplate code to pass the instance to the form, and will save the form.

You work with a Form , but a Form has no .instance attribute. A ModelForm has, so it might be better to use a ModelForm here:

class LibraryReserveForm(CrispyFormMixin, forms.ModelForm):

    def __init__(self, *args, **kwargs):
        self.manager = kwargs.pop('manager')
        super(LibraryReserveForm, self).__init__(*args, **kwargs)

    def save(self, *args, **kwargs):
        self.instance.reserve_status = 'approved'
        return super().save(*args, **kwargs)

Then we can make use of an UpdateView :

from django.views.generic import 

class LibraryReserveView(
          ProfileTypeRequiredMixin,
          MultiplePermissionsRequiredMixin,
          
    ):

    model = ReservedBooks
    template_name = 'library/reserved_list.html'
    form_class = LibraryReserveForm
    success_url = …

    def get_form_kwargs(self):
        kwargs = super(LibraryReserveView, self).get_form_kwargs()
        kwargs.update(
            manager=self.request.user.manager
        )
        return kwargs

You still need to specify the sucess_url here: the URL to which a successful POST request will redirect to implement the Post/Redirect/Get pattern [wiki] .