Initialize user django so he has default settings values - django

Question

I am using Django + Angular 2
. I am creating a user fine but he does not have the default values i want...
Model
class Settings(models.Model):
user = models.ForeignKey('auth.User', related_name='settings', on_delete=models.CASCADE)
bolean1 = models.BooleanField(default=False)
boolean2 = models.BooleanField(default=False)
boolean3 = models.BooleanField(default=False)
string1 = models.CharField(max_length=100, default='No description')
Serializer
class SettingsSerializer(serializers.ModelSerializer):
class Meta:
model = Settings
fields = ('id', 'bolean1', 'bolean1', 'bolean3', 'string1')
class UserSerializer(serializers.ModelSerializer):
class Meta:
model = User
fields = ('id', 'username', 'email', 'password' ,'settings', 'image')
extra_kwargs = {'password': {'write_only': True}}
class CreateUserSerializer(serializers.ModelSerializer):
class Meta:
model = User
fields = ('email', 'username', 'password')
extra_kwargs = {'password': {'write_only': True}}
def create(self, validated_data):
user = User(
email=validated_data['email'],
username=validated_data['username']
)
user.set_password(validated_data['password'])
user.save()
return user
Views
class SettingsValues(generics.ListAPIView):
serializer_class = SettingsSerializer
permission_classes = (permissions.IsAuthenticatedOrReadOnly,)
def get_queryset(self):
queryset = Settings.objects.all()
queryset = queryset.filter(user=self.request.user.id)
return queryset
class RegisterUser(generics.ListCreateAPIView):
queryset = User.objects.all()
serializer_class = CreateUserSerializer
The problem is that when i create a new user he does not have default values,e.g boolean 1, boolean 2 etc. i must go to django admin create new setting and choose the user.
Any idea ?

Answer 1

Sounds like you want Django's Signals, specifically post_save. You'll want to create a Settings() for each User() that's created - you have default values right now, but that only helps when you've saved the Settings model.
from django.db.models.signals import post_save
from django.dispatch import receiver
from django.contrib.auth import get_user_model
User = get_user_model()
#receiver(post_save, sender=User)
def create_settings(sender, instance, created, **kwargs):
# Only create a Settings relation if the User is new
if created:
settings = Settings(user=instance)
settings.save()
If your Settings model is in an app called settings, then your settings/apps.py should have this - good StackOverflow answer on this:
from django.apps import AppConfig
class SettingsConfig(AppConfig):
name = 'settings'
def ready(self):
from . import signals
Now, whenever a new User is created, a Settings relation is created for them as well.

Answer 2

I am not sure if this is what you want to achieve, but try this.
//admin.py
from django.contrib import admin
from django.contrib.auth.models import User
from django.contrib.auth.admin import UserAdmin
from accounts import models
admin.site.unregister(User)
class UserRoleInline(admin.StackedInline):
model = models.UserRole
class UserRoleAdmin(UserAdmin):
inlines = [UserRoleInline]
admin.site.register(User, UserRoleAdmin)
//model.py
from django.db import models
from django.contrib.auth.models import User
class RoleType(object):
MANAGER = 'm'
CANDIDATE = 'c'
CHOICES = ((MANAGER, 'Manager'), (CANDIDATE, 'Candidate'))
class UserRole(models.Model):
user = models.OneToOneField(User)
role = models.CharField(max_length=1, choices=RoleType.CHOICES)
Reference: django StackedInline
This will help you stack the setting models at end of User model. This won't automatically take a default value, but will let you choose the settings as the user models is created.