Found a better solution: class RequiredFormSet(BaseFormSet): def __init__(self, *args, **kwargs): super(RequiredFormSet, self).__init__(*args, **kwargs) for form in self.forms: form.empty_permitted = False Then create your formset like this: MyFormSet = formset_factory(MyForm, formset=RequiredFormSet) I really don’t know why this wasn’t an option to begin with… but, whatever. It only took a few hours of my life to figure … Read more
You can setup a custom ModelChoiceField that will return whatever label you’d like. Place something like this within a fields.py or wherever applicable. class UserModelChoiceField(ModelChoiceField): def label_from_instance(self, obj): return obj.get_full_name() Then when creating your form, simply use that field UserModelChoiceField(queryset=User.objects.filter(is_staff=False), required = False) More info can be found here
You don’t need to create any custom field, Django already does the job, just pass the choices well formatted: MEDIA_CHOICES = ( (‘Audio’, ( (‘vinyl’, ‘Vinyl’), (‘cd’, ‘CD’), ) ), (‘Video’, ( (‘vhs’, ‘VHS Tape’), (‘dvd’, ‘DVD’), ) ), )
model_instance = form.save(commit=False) will return you a object of the model without saving to the DB you can then add value of some field which is not available on form model_instance.some_field = value model_instance.save()
Yes. This is very much possible. You can read up on Form objects. It would be the same way you would treat a ModelForm, except that you are not bound by the model, and you have to explicitly declare all the form attributes. def form_handle(request): form = MyForm() if request.method==’POST’: form = MyForm(request.POST) if form.is_valid(): … Read more
For setting session variable: request.session[‘idempresa’] = profile.idempresa For getting session Data: if ‘idempresa’ in request.session: idempresa = request.session[‘idempresa’]
You can also use quite elegant add_error() method. Works for Django >= 1.7. If you set field as None form will treat the error as “non_field” one. So: form.add_error(None, “I’m non-field error”) works like a charm.
the first argument to your view should be request
I would recommend a custom Widget approach, HTML5 allows you to have a free text input with a dropdown list which would work as a pick-one-or-write-other type of field, this is how I made it: fields.py from django import forms class ListTextWidget(forms.TextInput): def __init__(self, data_list, name, *args, **kwargs): super(ListTextWidget, self).__init__(*args, **kwargs) self._name = name self._list … Read more
In regards to displaying a foreign key field in a form you can use the forms.ModelChoiceField and pass it a queryset. so, forms.py: class DocumentForm(forms.ModelForm): class Meta: model = Document def __init__(self, *args, **kwargs): user = kwargs.pop(‘user’,”) super(DocumentForm, self).__init__(*args, **kwargs) self.fields[‘user_defined_code’]=forms.ModelChoiceField(queryset=UserDefinedCode.objects.filter(owner=user)) views.py: def someview(request): if request.method==’post’: form=DocumentForm(request.POST, user=request.user) if form.is_valid(): selected_user_defined_code = form.cleaned_data.get(‘user_defined_code’) #do stuff … Read more