If I use normal user provided django default, my model will be like this.
from django.db import models
from django.contrib.auth.models import Users
class Photo(models.Model):
photographer=models.ForeignKey(User, on_delete=models.CASCADE, related_name='user') # 5
# other fields
If I want to implement allauth, what should I have to pass for the first argument in line#5, instead of User?
Django-allauth still uses the User model from django's auth, so you don't have to change anything. Unless you've extended a custom user model.
How come to import User from the auth package that comes with Django, I need to do:
from django.contrib.auth.models import User
while to refer to the same User model to create a ForeignKey, I need to do:
owner = models.ForeignKey('auth.User', related_name='snippets')
and not 'auth.models.User'?
I am following the example here : http://www.django-rest-framework.org/tutorial/4-authentication-and-permissions
This is due to Django's "lazy relationships". You can see the code for this here. You don't need to specify the exact module, in this case models because anything inheriting from Django's models.Model will fire off a class_prepared signal once it's initialised and up to this point, it's still only a string.
Just provide the app and model, or just the model name if it's in the same app.
I'm trying to make sure that the first name and last name field are not optional for the auth User model but I'm not sure how to change it. I can't use a sub class as I have to use the authentication system.
Two solutions I can think of are:
to put the name in the user profile but it's a little silly to have a field that I can't use correctly.
To validate in the form rather than in the model. I don't think this really fits with Django's philosophy...
For some reason I can't seem to find a way to do this online so any help is appreciated. I would have thought that this would be a popular question.
Cheers,
Durand
Simplest solution
Just create a custom UserRegisterForm which inherits the django's default UserCreationForm.
The first_name and last_name are already attributes of django's default User. If you want to make them as required fields, then recreate those fields as forms.CharField(...).
Now use your own User register form.
# Contents usersapp/forms.py
from django import forms
from django.contrib.auth.models import User
from django.contrib.auth.forms import UserCreationForm
# Inherit Django's default UserCreationForm
class UserRegisterForm(UserCreationForm):
first_name = forms.CharField(max_length=50) # Required
last_name = forms.CharField(max_length=50) # Required
# All fields you re-define here will become required fields in the form
class Meta:
model = User
fields = ['username', 'email', 'first_name', 'last_name', 'password1', 'password2']
I would definitely go with validating on the form. You could even go as far as having more form validation in the admin if you felt like it.
Thanks Mbuso for the advice. Here's my full implementation for those who are interested. Before taking a look at the source, let's see what it looks like:
I've implemented a profile model, but this will work just fine without it.
from django.core.exceptions import ValidationError
from django.contrib import admin
from django.contrib.auth.admin import UserAdmin
from django.contrib.auth.forms import UserChangeForm
from django.contrib.auth.models import User
from apps.profiles.models import Profile
# Define an inline admin descriptor for Profile model
# which acts a bit like a singleton
class UserProfileInline(admin.StackedInline):
model = Profile
can_delete = False
verbose_name_plural = 'profile'
class MyUserChangeForm(UserChangeForm):
def clean_first_name(self):
if self.cleaned_data["first_name"].strip() == '':
raise ValidationError("First name is required.")
return self.cleaned_data["first_name"]
def clean_last_name(self):
if self.cleaned_data["last_name"].strip() == '':
raise ValidationError("Last name is required.")
return self.cleaned_data["last_name"]
# Define a new User admin
class MyUserAdmin(UserAdmin):
form = MyUserChangeForm
inlines = UserProfileInline,
admin.site.unregister(User)
admin.site.register(User, MyUserAdmin)
Note: If you do implement a profile model, recommend using UserProfile as the name, since is this is what's in the documentation and seems to be the standard (this part was developed before I started working on the project). If you're using Django 1.5 or higher, skip UserProfile all together and extend the User model.
The Django way of extending the basic User model is through user profiles: see "Storing additional information about users".
If it does not fit your needs, django.contrib.auth is just a Django application, I would simply fork it. As long as you abide by the original interface, I think you will be out of trouble.
Another option is Pinax - it has OpenId support built in, you can use it with your own openid provider. OpenId native support is a battery I really miss in Django.
I'm learning Python Django framework. Thanks to StackOverFlow Community, i have learned how to use open-id in django (social_auth i'm using ).
with social_auth, sign in and sign out process are pretty easy and practical. But i want to build friendship relationship between users, for this i created new model class which is extended from from django.contrib.auth.models import User and add control when new user is created in auth_table, i create new user on CustomUser Table.(It is not efficient because the parameters coming from open-id are different for each open-id connection)
My model code is here
from datetime import datetime
from django.core.exceptions import ObjectDoesNotExist
from django.db import models
from django.contrib.auth.models import User
from django.contrib.auth.models import UserManager
# Create your models here.
class CustomUser(User):
gender = models.CharField(max_length=50)
location = models.CharField(max_length=50)
bio = models.CharField(max_length=50)
web_pages = models.CharField(max_length=50)
friends = models.ManyToManyField("self")
And How can i use CustomUser model and OpenID together in Django for authentication?
Any Help will be appreciated.
The way to extend the User class in Django is to provide user profile class. See here. You can then just use the User model as always and get the profile instance on demand.
What is the best approach to extending the Site model in django? Creating a new model and ForeignKey the Site or there another approach that allows me to subclass the Site model?
I prefer subclassing, because relationally I'm more comfortable, but I'm concerned for the impact it will have with the built-in Admin.
I just used my own subclass of Site and created a custom admin for it.
Basically, when you subclass a model in django it creates FK pointing to parent model and allows to access parent model's fields transparently- the same way you'd access parent class attributes in pyhon.
Built in admin won't suffer in any way, but you'll have to un-register Sites ModelAdmin and register your own ModelAdmin.
If you only want to change behaviour of the object, but not add any new fields, you should consider using a "proxy model" (new in Django 1.1). You can add extra Python methods to existing models, and more:
This is what proxy model inheritance is for: creating a proxy for the original model. You can create, delete and update instances of the proxy model and all the data will be saved as if you were using the original (non-proxied) model. The difference is that you can change things like the default model ordering or the default manager in the proxy, without having to alter the original.
Read more in the documentation.
As of Django 2.2 there still no simple straight way to extend Site as can be done for User. Best way to do it now is to create new entity and put parameters there. This is the only way if you want to leverage existing sites support.
class SiteProfile(models.Model):
title = models.TextField()
site = models.OneToOneField(Site, on_delete=models.CASCADE)
You will have to create admin for SiteProfile. Then add some SiteProfile records with linked Site. Now you can use site.siteprofile.title anywhere where you have access to current site from model.
You can have another model like SiteProfile which has a OneToOne relation with Site.
It has been a long time since the question was asked, but I think there is not yet (Django 3.1) an easy solution for it like creating a custom user model. In this case, creating a custom user model inheriting from django.contrib.auth.models.AbstractUser model and changing AUTH_USER_MODEL (in settings) to the newly created custom user model solves the issue.
However, it can be achieved for also Site model with a long solution written below:
SOLUTION
Suppose that you have an app with the name core. Use that app for all of the code below, except the settings file.
Create a SiteProfile model with a site field having an OneToOne relation with the Site model. I have also changed its app_label meta so it will be seen under the Sites app in the admin.
# in core.models
...
from django.contrib.sites.models import Site
from django.db import models
class SiteProfile(models.Model):
"""SiteProfile model is OneToOne related to Site model."""
site = models.OneToOneField(
Site, on_delete=models.CASCADE, primary_key=True,
related_name='profiles', verbose_name='site')
long_name = models.CharField(
max_length=255, blank=True, null=True)
meta_name = models.CharField(
max_length=255, blank=True, null=True)
def __str__(self):
return self.site.name
class Meta:
app_label = 'sites' # make it under sites app (in admin)
...
Register the model in the admin. (in core.admin)
What we did until now was good enough if you just want to create a site profile model. However, you will want the first profile to be created just after migration. Because the first site is created, but not the first profile related to it. If you don't want to create it by hand, you need the 3rd step.
Write below code in core.apps.py:
# in core.apps
...
from django.conf import settings
from django.db.models.signals import post_migrate
def create_default_site_profile(sender, **kwargs):
"""after migrations"""
from django.contrib.sites.models import Site
from core.models import SiteProfile
site = Site.objects.get(id=getattr(settings, 'SITE_ID', 1))
if not SiteProfile.objects.exists():
SiteProfile.objects.create(site=site)
class CoreConfig(AppConfig):
name = 'core'
def ready(self):
post_migrate.connect(create_default_site_profile, sender=self)
from .signals import (create_site_profile) # now create the second signal
The function (create_default_site_profile) will automatically create the first profile related to the first site after migration, using the post_migrate signal. However, you will need another signal (post_save), the last row of the above code.
If you do this step, your SiteProfile model will have a full connection with the Site model. A SiteProfile object is automatically created/updated when any Site object is created/updated. The signal is called from apps.py with the last row.
# in core.signals
from django.contrib.sites.models import Site
from django.db.models.signals import post_save, post_migrate
from django.dispatch import receiver
from .models import SiteProfile
#receiver(post_save, sender=Site)
def create_site_profile(sender, instance, **kwargs):
"""This signal creates/updates a SiteProfile object
after creating/updating a Site object.
"""
siteprofile, created = SiteProfile.objects.update_or_create(
site=instance
)
if not created:
siteprofile.save()
Would you like to use it on templates? e.g.
{{ site.name }}
Then you need the 5th and 6th steps.
Add the below code in settings.py > TEMPLATES > OPTIONS > context_processors
'core.context_processors.site_processor'
# in settings.py
TEMPLATES = [
{
# ...
'OPTIONS': {
'context_processors': [
# ...
# custom processor for getting the current site
'core.context_processors.site_processor',
],
},
},
]
Create a context_processors.py file in the core app with the code below.
A try-catch block is needed (catch part) to make it safer. If you delete all sites from the database you will have an error both in admin and on the front end pages. Error is Site matching query does not exist. So the catch block creates one if it is empty.
This solution may not be fully qualified if you have a second site and it is deleted. This solution only creates a site with id=1.
# in core.context_processors
from django.conf import settings
from django.contrib.sites.models import Site
def site_processor(request):
try:
return {
'site': Site.objects.get_current()
}
except:
Site.objects.create(
id=getattr(settings, 'SITE_ID', 1),
domain='example.com', name='example.com')
You can now use the site name, domain, meta_name, long_name, or any field you added, in your templates.
# e.g.
{{ site.name }}
{{ site.profiles.long_name }}
It normally adds two DB queries, one for File.objects and one for FileProfile.objects. However, as it is mentioned in the docs,
Django is clever enough to cache the current site at the first request and it serves the cached data at the subsequent calls.
https://docs.djangoproject.com/en/3.1/ref/contrib/sites/#caching-the-current-site-object
Apparently, you can also create a models.py file in a folder that you add to INSTALLED_APPS, with the following content:
from django.contrib.sites.models import Site as DjangoSite, SiteManager
from django.core.exceptions import ImproperlyConfigured
from django.db import models
from django.http.request import split_domain_port
# our site model
class Site(DjangoSite):
settings = models.JSONField(blank=True, default={})
port = models.PositiveIntegerField(null=True)
protocol = models.CharField(default='http', max_length=5)
#property
def url(self):
if self.port:
host = f'{self.domain}:{self.port}'
else:
host = self.domain
return f'{self.protocol}://{host}/'
# patch django.contrib.sites.models.Site.objects to use our Site class
DjangoSite.objects.model = Site
# optionnal: override get_current to auto create site instances
old_get_current = SiteManager.get_current
def get_current(self, request=None):
try:
return old_get_current(self, request)
except (ImproperlyConfigured, Site.DoesNotExist):
if not request:
return Site(domain='localhost', name='localhost')
host = request.get_host()
domain, port = split_domain_port(host)
Site.objects.create(
name=domain.capitalize(),
domain=host,
port=port,
protocol=request.META['wsgi.url_scheme'],
)
return old_get_current(self, request)
SiteManager.get_current = get_current
In my opinion, the best way to doing this is by writing a model related to the site model using inheritance
First, add the site id to the Django settings file
SITE_ID = 1
now create a model in one of your apps
from django.db import models
from django.contrib.sites.models import Site
class Settings(Site):
field_a = models.CharField(max_length=150, null=True)
field_b = models.CharField(max_length=150, null=True)
class Meta:
verbose_name_plural = 'settings'
db_table = 'core_settings' # core is name of my app
def __str__(self) -> str:
return 'Settings'
then edit the apps.py file of that app
from django.apps import AppConfig
from django.db.models.signals import post_migrate
def build_settings(sender, **kwargs):
from django.contrib.sites.models import Site
from .models import Settings
if Settings.objects.count() < 1:
Settings.objects.create(site_ptr=Site.objects.first())
class CoreConfig(AppConfig):
default_auto_field = 'django.db.models.BigAutoField'
name = 'project.apps.core'
def ready(self) -> None:
post_migrate.connect(build_settings, sender=self)
now every time you run migrations a row will be auto-generated in core_settings that have a one to one relationship with your Site model
and now you can access your settings like this
Site.objects.get_current().settings.access_id
optional: if have only one site
unregister site model from admin site and disable deleting and creating settings model in admin site
from django.contrib import admin
from . import models
from django.contrib.sites.models import Site
admin.site.unregister(Site)
#admin.register(models.Settings)
class SettingAdminModel(admin.ModelAdmin):
def has_delete_permission(self, request,obj=None) -> bool:
return False
def has_add_permission(self, request) -> bool:
return False