How can a ChoiceField.choices callable know what choices to return?

Question

In Django 1.8, the ChoiceField's choices argument can accept a callable:

def get_choices():
    return [(1, "one"), (2, "two")]

class MyForm(forms.Form):
    my_choice_field = forms.ChoiceField(choices=get_choices)

In the above example, get_choices() always returns the same choices. However, being able to assign a callable to choices does not make much sense unless that callable knows something like, say, an object id, each time it is called. How can I pass such a thing to it?

Answer 1

You can't do it in the form declaration because the CallableChoiceIterator calls the function without arguments that he gets from here. Doing in the __init__ Form method is easier than creating your own ChoiceField I guess. Here is what I suggest:

class MyForm(forms.Form):
    my_choice_field = forms.ChoiceField(choices=())

    def __init__(self, *args, **kwargs):
        # Let's pass the object id as a form kwarg
        self.object_id = kwargs.pop('object_id') 

        # django metaclass magic to construct fields
        super().__init__(*args, **kwargs)

        # Now you can get your choices based on that object id            
        self.fields['my_choice_field'].choices = your_get_choices_function(self.object_id)

That supposes that you have some Class Based View that looks that has a method like this :

class MyFormView(FormView):
   # ...

   def get_form_kwargs(self):
       kwargs = super().get_form_kwargs()
       kwargs['object_id'] = 'YOUR_OBJECT_ID_HERE'
       return kwargs

   # ...

P.S : The super() function call supposes you are using python 3