Django admin choice field - django

Question

I have a model that has a CharField and in the admin I want to add choices to the widget. The reason for this is I'm using a proxy model and there are a bunch of models that share this CharField but they each have different choices.
class MyModel(MyBaseModel):
stuff = models.CharField('Stuff', max_length=255, default=None)
class Meta:
proxy = True
class MyModelAdmin(admin.ModelAdmin):
fields = ('stuff',)
list_display = ('stuff',)
admin.site.register(MyModel, MyModelAdmin)
For this model I want to use MY_CHOICES in MyModelAdmin.
Do I override a widget? Do I need to override the whole form?

Answer 1

from django.contrib import admin
from django import forms
class MyModel(MyBaseModel):
stuff = models.CharField('Stuff', max_length=255, default=None)
class Meta:
proxy = True
class MyModelForm(forms.ModelForm):
MY_CHOICES = (
('A', 'Choice A'),
('B', 'Choice B'),
)
stuff = forms.ChoiceField(choices=MY_CHOICES)
class MyModelAdmin(admin.ModelAdmin):
fields = ('stuff',)
list_display = ('stuff',)
form = MyModelForm
admin.site.register(MyModel, MyModelAdmin)
See: https://docs.djangoproject.com/en/dev/ref/forms/fields/#choicefield

Answer 2

You don't need a custom form.
This is the minimum you need:
# models.py
from __future__ import unicode_literals
from django.db import models
class Photo(models.Model):
CHOICES = (
('hero', 'Hero'),
('story', 'Our Story'),
)
name = models.CharField(max_length=250, null=False, choices=CHOICES)
# admin.py
from django.contrib import admin
from .models import Photo
class PhotoAdmin(admin.ModelAdmin):
list_display = ('name',)
admin.site.register(Photo, PhotoAdmin)

Answer 3

You can override formfield_for_choice_field() that way you don't need to create a new form.
class MyModelAdmin(admin.ModelAdmin):
def formfield_for_choice_field(self, db_field, request, **kwargs):
if db_field.name == 'status':
kwargs['choices'] = (
('accepted', 'Accepted'),
('denied', 'Denied'),
)
if request.user.is_superuser:
kwargs['choices'] += (('ready', 'Ready for deployment'),)
return super().formfield_for_choice_field(db_field, request, **kwargs)
See formfield_for_choice_field

Answer 4

You need to override the form the ModelAdmin is going to use:
class MyForm(forms.ModelForm):
stuff = forms.CharField('Stuff', max_length=255, choices=MY_CHOICES, default=None)
class Meta:
model = MyModel
fields = ('stuff', 'other_field', 'another_field')
class MyModelAdmin(admin.ModelAdmin):
fields = ('stuff',)
list_display = ('stuff',)
form = MyForm
If you need your choices to be dynamic, maybe you could do something similar to:
class MyForm(forms.ModelForm):
stuff = forms.CharField('Stuff', max_length=255, choices=MY_CHOICES, default=None)
def __init__(self, stuff_choices=(), *args, **kwargs):
# receive a tupple/list for custom choices
super(MyForm, self).__init__(*args, **kwargs)
self.fields['stuff'].choices = stuff_choices
and in your ModelAdmin's __init__ define what MY_CHOICES is going to be and assign the form instance there instead:
Good luck! :)

Answer 5

in Gerard's answer, if you keep :
def __init__(self, stuff_choices=(), *args, **kwargs):
then when you will try to add new model from admin, you will always get 'This field is required.' for all required fields.
you should remove stuff_choices=() from initialization:
def __init__(self,*args, **kwargs):

Answer 6

You need to think of how you are going to store the data at a database level.
I suggest doing this:
Run this pip command: pip install django-multiselectfield
In your models.py file:
from multiselectfield import MultiSelectField
MY_CHOICES = (('item_key1', 'Item title 1.1'),
('item_key2', 'Item title 1.2'),
('item_key3', 'Item title 1.3'),
('item_key4', 'Item title 1.4'),
('item_key5', 'Item title 1.5'))
class MyModel(models.Model):
my_field = MultiSelectField(choices=MY_CHOICES)
In your settings.py:
INSTALLED_APPS = (
'django.contrib.auth',
'django.contrib.contenttypes',
'django.contrib.sessions',
'django.contrib.sites',
'django.contrib.admin',
#.....................#
'multiselectfield',
)
Watch the MAGIC happen!
Source:
https://pypi.python.org/pypi/django-multiselectfield

Answer 7

Below a solution that works immediately with Postgres' special ArrayField:
# models.py
class MyModel(models.Model):
class Meta:
app_label = 'appname'
name = models.CharField(max_length=1000, blank=True)
ROLE_1 = 'r1'
ROLE_2 = 'r2'
ROLE_3 = 'r3'
ROLE_CHOICES = (
(ROLE_1, 'role 1 name'),
(ROLE_2, 'role 2 name'),
(ROLE_3, 'role 3 name'),
)
roles = ArrayField(
models.CharField(choices=ROLE_CHOICES, max_length=2, blank=True),
default=list
)
# admin.py
class MyModelForm(ModelForm):
roles = MultipleChoiceField(choices=MyModel.ROLE_CHOICES, widget=CheckboxSelectMultiple)
#admin.register(MyModel)
class MyModelAdmin(admin.ModelAdmin):
form = MyModelForm
list_display = ("pk", "name", "roles")
(Django 2.2)