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Specific Queryset for Input on Django ModelAdmin Change Form

I've got a 'Registration' object in place that users can create on the front end without issue.
It looks like this:
class Registration(models.Model):
person = models.ForeignKey(Person, on_delete=models.PROTECT)
course_detail = models.ForeignKey(CourseDetail, on_delete=models.PROTECT)
camp_shirt = models.ForeignKey(CampShirt, on_delete=models.PROTECT)
comments = models.CharField(max_length=200, blank=True, null=True)
created_at = models.DateTimeField(auto_now_add=True)
updated_at = models.DateTimeField(auto_now=True)
def __str__(self):
return "%s" % (self.course_detail.course.camp)
When I am in the admin and click on a given Registration - it takes a while to load because there are thousands and thousands of Person objects.
For ease of use - there will never be a time when we would need to edit the 'person' associated with a given registration, so I would like to make the 'person' dropdown only show the selected user in the person queryset when editing from the django admin.
So when I go to http://myapp.com/admin/registration/23/change I want the form to only display the currently selected person as the only option in the dropdown.
My admin model looks like this:
class RegistrationAdmin(admin.ModelAdmin):
list_display = ("person", "course_detail")
class Meta:
# I think this is what I do in order to override the default admin form? Not sure.
form = RegistrationAdminForm
My RegistrationAdminForm looks like this:
class RegistrationAdminForm(forms.ModelForm):
# course_detail, person, camp_shirt, comments
person = forms.ModelChoiceField(queryset=Person.objects.filter(
id=registration.person.id)
)
def __init__(self, registration, *args, **kwargs):
super(RegistrationAdminForm, self).__init__(*args, **kwargs)
self.fields['person'].queryset = Person.objects.filter(
id=registration.person.id
)
class Meta:
model = Registration
fields = '__all__'
Main Question : How do I change the admin form so that a specific queryset is returned for one of the fields in the django admin?

If the person field will never be changed you can add the person field to readonly_fields, a select with all Person objects will not be rendered.
class RegistrationAdmin(admin.ModelAdmin):
list_display = ("person", "course_detail")
readonly_fields = ("person", )
Then you do not need your custom form. FYI when you want to add a custom form to a ModelAdmin you do not put it in Meta, you define it on the form itself
class MyModelAdmin(admin.ModelAdmin):
form = MyModelForm

remove the default select in ForeignKey Field of django admin

There are 150k entries in User model. When i am using it in django-admin without the raw_id_fields it is causing problem while loading all the entries as a select menu of foreign key. is there alternate way so that it could be loaded easily or could become searchable?
I have these models as of defined above and there is a User model which is used as ForeignKey in ProfileRecommendation models. The database entry for user model consist of around 150k entries. I don't want default select option for these foreign fields. Instead if can filter them out and load only few entries of the user table. How I can make them searchable like autocomplete suggestion?
admin.py
class ProfileRecommendationAdmin(admin.ModelAdmin):
list_display = ('user', 'recommended_by', 'recommended_text')
raw_id_fields = ("user", 'recommended_by')
search_fields = ['user__username', 'recommended_by__username',]
admin.site.register(ProfileRecommendation, ProfileRecommendationAdmin)
models.py
class ProfileRecommendation(models.Model):
user = models.ForeignKey(User, related_name='recommendations')
recommended_by = models.ForeignKey(User, related_name='recommended')
recommended_on = models.DateTimeField(auto_now_add=True, null=True)
recommended_text = models.TextField(default='')

you can use method formfield_for_foreignkey
something like this:
class ProfileRecommendationAdmin(admin.ModelAdmin):
def formfield_for_foreignkey(self, db_field, request, **kwargs):
if db_field.name == "user":
kwargs["queryset"] = User.objects.filter(is_superuser=True)
return super(ProfileRecommendationAdmin,self).formfield_for_foreignkey(db_field, request, **kwargs)
or other way is to override the form for that modeladmin class
from django import forms
from django.contrib import admin
from myapp.models import Person
class PersonForm(forms.ModelForm):
class Meta:
model = Person
exclude = ['name']
class PersonAdmin(admin.ModelAdmin):
exclude = ['age']
form = PersonForm
you can change the widget and use something like the selectize with autocomplete and ajax.

Django rest framework displaying specific field of a model

In my model:
from django.contrib.auth.models import User
class Restaurant(models.Model):
manager = models.ForeignKey(User, on_delete=models.PROTECT,
null=True, blank=False, related_name="manager")
in my serializers.py
class RestaurantSerializer(CoreHyperlinkedModelSerializer):
class Meta:
model = Restaurant
in my views.py
class RestaurantViewSet(viewsets.ModelViewSet):
queryset = Restaurant.objects.order_by('id').all()
serializer_class = RestaurantSerializer
on my list:
the manager is displaying as <rest_framework.relations.PKOnlyObject object at 0x9f7040xbc208>
How can I display it as normal data like its username?

You want to use a 'SlugRelatedField'.

There are a few ways you can go, but if you just want to show a username, all you need is this
from rest_framework import serializers
class RestaurantSerializer(serializers.ModelSerializer):
manager = serializers.CharField(source="manager.username")
class Meta:
model = Restaurant
if you inherit from ModelSerializer and skip the manager field, it will use user PK as the value of the manager field by default.
a slightly more involved way would be to define a separate serializer for User and then embed it in RestaurantSerializer.
from rest_framework import serializers
class UserSerializer(serializers.ModelSerializer):
class Meta:
model = User
class RestaurantSerializer(serializers.ModelSerializer):
manager = UserSerializer()
class Meta:
model = Restaurant
And if you really want to use hyperlinked serializer, you need to do quite a bit of work. You need to read this part carefully http://www.django-rest-framework.org/api-guide/serializers/#how-hyperlinked-views-are-determined

Django multiple user profiles - Display user fields inline in admin

I am relatively new to Django and I'm trying to achieve something that is not quite clear in the documentation. My application requires multiple types of users. So, I have extended django User, creating a Profile model that contains additional fields common in all User types:
USER_TYPES = (
('A', 'UserTypeA'),
('B', 'UserTypeB'),
('C', 'UserTypeC'),
)
class Profile(models.Model):
user = models.OneToOneField(User, unique=True)
about = models.TextField(blank=True)
user_type = models.CharField(max_length=3, choices=USER_TYPES, default='A')
def save(self, *args, **kwargs):
try:
existing = Profile.objects.get(user=self.user)
self.id = existing.id #force update instead of insert
except Profile.DoesNotExist:
print "Profile not created yet"
models.Model.save(self, *args, **kwargs)
def create_user(sender, instance, created, **kwargs):
print "User Profile Creation: False"
if created:
print "User Profile Creation: ", created
Profile.objects.create(user=instance)
post_save.connect(create_user, sender=Profile)
In settings.py I have set:
AUTH_PROFILE_MODULE = 'users.Profile'
After that i have defined my UserTypeX models deriving from Profile models like this:
class UserTypeA(Profile):
phone_number = models.CharField(max_length=13,blank=False)
class UserTypeB(Profile):
company_name = models.CharField(max_length=30,blank=False)
phone_number = models.CharField(max_length=13,blank=False)
def __init__(self, *args, **kwargs):
kwargs['user_type'] = 'B'
super(UserTypeB, self).__init__(*args, **kwargs)
...
I've registered those user models to admin so that I could manage my users independently.
The default admin behavior displays correctly all the Profile and UserTypeX fields and there is a select box (with a plus button next to it) for the User field - due to the OneToOneField relationship between Profile and User models. So whenever I want to create a new UserTypeX, I have to press the plus button and fill in a new popup window all the default User django defined fields.
What I am struggling to do now is display those User fields, not in a new popup window but inline in my UserTypeX add/edit page. I've read the documentation about StackedInlines and TabularInlines but those doesn't fit my case because I want to inline parent fields in an ancestor's add/edit page and not vice versa.
Is there any suggested solution with example code (please!) for that problem? Thank's in advance!
So, to make things short, is there a way to display User fields (instead of the select/add functionality due to OneToOneField relationship) in Profile add/edit screen in admin?
Update: A related question (unanswered though...) that briefly addresses the problem is:
Reverse Inlines in Django Admin

No, you cannot. But you can make B inline in A if you want. Or maybe you can manipulate how django display it in
def unicode(self):
models.py
class A(models.Model):
name = models.CharField(max_length=50)
def __unicode__(self):
return self.name
class B(models.Model):
name = models.CharField(max_length=50)
a = models.ForeignKey(A)
admin.py
class B_Inline(admin.TabularInline):
model = B
class A_Admin(admin.ModelAdmin):
inlines = [
B_Inline,
]
admin.site.register(A, A_Admin)
admin.site.register(B)
Or maybe you want to use many-to-many relationship?
models.py
class C(models.Model):
name = models.CharField(max_length=50)
def __unicode__(self):
return self.name
class D(models.Model):
name = models.CharField(max_length=50)
cs = models.ManyToManyField(C)
admin.py
class C_Inline(admin.TabularInline):
model = D.cs.through
class D_Admin(admin.ModelAdmin):
exclude = ("cs",)
inlines = [
C_Inline,
]
admin.site.register(C)
admin.site.register(D, D_Admin)

I don't see how this is going to work - you have told django to use users.Profile as your profile model but you actually want to use it's children. Plus as you say you're wanting to use the inline forms the "wrong way around". Without knowing more about what exactly you're trying to achieve I'd suggest that you could fix the inline form issue by defining a custom view to edit the user details rather than using django admin. As for the different profile types - I'd suggest just defining a profile type field in the model and then hiding/showing different fields/properties depending on its value

Django Admin: OneToOne Relation as an Inline?

I'm putting together the admin for a satchmo application. Satchmo uses OneToOne relations to extend the base Product model, and I'd like to edit it all on one page.
Is it possible to have a OneToOne relation as an Inline? If not, what is the best way to add a few fields to a given page of my admin that will eventually be saved into the OneToOne relation?
for example:
class Product(models.Model):
name = models.CharField(max_length=100)
...
class MyProduct(models.Model):
product = models.OneToOne(Product)
...
I tried this for my admin but it does not work, and seems to expect a Foreign Key:
class ProductInline(admin.StackedInline):
model = Product
fields = ('name',)
class MyProductAdmin(admin.ModelAdmin):
inlines = (AlbumProductInline,)
admin.site.register(MyProduct, MyProductAdmin)
Which throws this error: <class 'satchmo.product.models.Product'> has no ForeignKey to <class 'my_app.models.MyProduct'>
Is the only way to do this a Custom Form?
edit: Just tried the following code to add the fields directly... also does not work:
class AlbumAdmin(admin.ModelAdmin):
fields = ('product__name',)

It's perfectly possible to use an inline for a OneToOne relationship. However, the actual field defining the relationship has to be on the inline model, not the parent one - in just the same way as for a ForeignKey. Switch it over and it will work.
Edit after comment: you say the parent model is already registered with the admin: then unregister it and re-register.
from original.satchmo.admin import ProductAdmin
class MyProductInline(admin.StackedInline):
model = MyProduct
class ExtendedProductAdmin(ProductAdmin):
inlines = ProductAdmin.inlines + (MyProductInline,)
admin.site.unregister(Product)
admin.site.register(Product, ExtendedProductAdmin)
Update 2020 (Django 3.1.1)
This method is still working but some types has changed in new Django version since inlines in ExtendedProductAdmin should now be added as list and not tuple, like this:
class ExtendedProductAdmin(ProductAdmin):
inlines = ProductAdmin.inlines + [MyProductInline]
Or you will get this error:
inlines = ProductAdmin.inlines + (MyProductInline,)
TypeError: can only concatenate list (not "tuple") to list

Maybe use inheritance instead OneToOne relationship
class Product(models.Model):
name = models.CharField(max_length=100)
...
class MyProduct(Product):
.....
Or use proxy classes
class ProductProxy(Product)
class Meta:
proxy = True
in admin.py
class MyProductInlines(admin.StackedInline):
model = MyProduct
class MyProductAdmin(admin.ModelAdmin):
inlines = [MyProductInlines]
def queryset(self, request):
qs = super(MyProductAdmin, self).queryset(request)
qs = qs.exclude(relatedNameForYourProduct__isnone=True)
return qs
admin.site.register(ProductProxy, MyProductAdmin)
In this variant your product will be in inline.

Referring to the last question, what would be the best solution for multiple sub-types. E.g class Product with sub-type class Book and sub-type class CD. The way shown here you would have to edit a product the general items plus the sub-type items for book AND the sub-type items for CD. So even if you only want to add a book you also get the fields for CD. If you add a sub-type e.g. DVD, you get three sub-type field groups, while you actually only want one sub-type group, in the mentioned example: books.

You can also try setting 'parent_link=True' on your OneToOneField?
https://docs.djangoproject.com/en/dev/topics/db/models/#specifying-the-parent-link-field

Jun, 2022 Update:
Yes, it's possible to have inline for one-to-one relation.
For example, as shown below, if "MyProduct" class has "models.OneToOneField()" referring to "Product" class which means "MyProduct" class has the ForeignKey referring to "Product" class:
# "models.py"
from django.db import models
class Product(models.Model):
name = models.CharField(max_length=100)
class MyProduct(models.Model):
name = models.CharField(max_length=100)
product = models.OneToOneField( # Here
Product,
on_delete=models.CASCADE,
primary_key=True
)
Then, you can inline "MyProduct" class under "Product" class as shown below:
# "admin.py"
from django.contrib import admin
from .models import Product, MyProduct
class MyProductInline(admin.TabularInline):
model = MyProduct
#admin.register(Product)
class ProductAdmin(admin.ModelAdmin):
inlines = (MyProductInline, )
Oppositely, as shown below, if "Product" class has "models.OneToOneField()" referring to "MyProduct" class which means "Product" class has the ForeignKey referring to "MyProduct" class:
# "models.py"
from django.db import models
class MyProduct(models.Model):
name = models.CharField(max_length=100)
class Product(models.Model):
name = models.CharField(max_length=100)
my_product = models.OneToOneField( # Here
MyProduct,
on_delete=models.CASCADE,
primary_key=True
)
Then, you can inline "Product" class under "MyProduct" class as shown below:
# "admin.py"
from django.contrib import admin
from .models import Product, MyProduct
class ProductInline(admin.TabularInline):
model = Product
#admin.register(MyProduct)
class MyProductAdmin(admin.ModelAdmin):
inlines = (ProductInline, )