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Specify Django proxy model in admin foreign key

In Django 1.11, I have a model Friend, and a proxy model Relative:
class FriendManager(models.Manager):
def get_queryset(self):
return super(RelativeManager, self).get_queryset().filter(is_relative=False)
class Friend(models.Model):
# Model fields defined here
objects = FriendManager()
class RelativeManager(models.Manager):
def get_queryset(self):
return super(RelativeManager, self).get_queryset().filter(is_relative=True)
class Relative(Friend):
class Meta:
proxy = True
objects = RelativeManager()
def save(self, *args, **kwargs):
self.is_relative = True
super(Relative, self).save(*args, **kwargs)
I also have a model FriendPortrait, which has a foreign key field friend:
class FriendPortrait(models.Model):
friend = models.ForeignKey(Friend)
And a proxy on that:
class RelativePortrait(FriendPortrait):
class Meta:
proxy = True
Now, I want the detail view for RelativePortraits to only show relatives in the drop-down for friend.
admin.py:
#admin.register(RelativePortrait)
class RelativePortraitAdmin(admin.ModelAdmin):
fields = ('friend')
def formfield_for_foreignkey(self, db_field, request, **kwargs):
if db_field.name == 'friend':
kwargs['queryset'] = Relative.objects.all()
return super(RelativePortraitAdmin, self).formfield_for_foreignkey(db_field, request, **kwargs)
This works, in that only relatives are displayed in the friend drop-down. However, when I try to save a portrait, Django admin gives me a validation error:
friend instance with id 14 does not exist.
How can I specify that I want to use a proxy model for my foreign key in the RelativePortraitAdmin?

The problem here is that your ForeignKey points to the Friend model. The model's default manager filters out all relatives, so this will not work.
A simple way to solve this would be to restructure your models a bit. Introducing something like a generic Person model and having Friend and Relative inherit from it with proxy=True. The Person model shouldn't have a manager that pre-filters the instances; then you could have your ForeignKey point to person.

Limit choices and validate django's foreign key to related objects (also in REST)

I have my models.py like this:
class Category(models.Model):
user = models.ForeignKey(User)
name = models.CharField(max_length=256, db_index=True)
class Todo(models.Model):
user = models.ForeignKey(User)
category = models.ForeignKey(Category)
...
And I want to limit choices of Category for Todo to only those ones where Todo.user = Category.user
Every solutuion that I've found was to set queryset for a ModelForm or implement method inside a form. (As with limit_choices_to it is not possible(?))
The problem is that I have not only one model with such limiting problem (e.g Tag, etc.)
Also, I'm using django REST framework, so I have to check Category when Todo is added or edited.
So, I also need functions validate in serializers to limit models right (as it does not call model's clean, full_clean methods and does not check limit_choices_to)
So, I'm looking for a simple solution, which will work for both django Admin and REST framework.
Or, if it is not possible to implement it the simple way, I'm looking for an advice of how to code it the most painless way.

Here what I've found so far:
To get Foreignkey showed right in admin, you have to specify a form in ModelAdmin
class TodoAdminForm(ModelForm):
def __init__(self, *args, **kwargs):
super().__init__(*args, **kwargs)
self.fields['category'].queryset = Category.objects.filter(user__pk=self.instance.user.pk)
#admin.register(Todo)
class TodoAdmin(admin.ModelAdmin):
form = TodoAdminForm
...
To get ManyToManyField showed right in InlineModelAdmin (e.g. TabularInline) here comes more dirty hack (can it be done better?)
You have to save your quiring field value from object and then manually set queryset in the field. My through model has two members todo and tag
And I'd like to filter tag field (pointing to model Tag):
class MembershipInline(admin.TabularInline):
model = Todo.tags.through
def get_formset(self, request, obj=None, **kwargs):
request.saved_user_pk = obj.user.pk # Not sure if it can be None
return super().get_formset(request, obj, **kwargs)
def formfield_for_foreignkey(self, db_field, request=None, **kwargs):
if db_field.name == 'tag':
kwargs['queryset'] = Tag.objects.filter(user__pk=request.saved_user_pk)
return super().formfield_for_foreignkey(db_field, request, **kwargs)
And finally, to restrict elements only to related in Django REST framework, I have to implement custom Field
class PrimaryKeyRelatedByUser(serializers.PrimaryKeyRelatedField):
def get_queryset(self):
return super().get_queryset().filter(user=self.context['request'].user)
And use it in my serializer like
class TodoSerializer(serializers.ModelSerializer):
category = PrimaryKeyRelatedByUser(required=False, allow_null=True, queryset=Category.objects.all())
tags = PrimaryKeyRelatedByUser(required=False, many=True, queryset=Tag.objects.all())
class Meta:
model = Todo
fields = ('id', 'category', 'tags', ...)
Not sure if it actually working in all cases as planned. I'll continue this small investigation.
Question still remains. Could it be done simplier?

How to delete one-to-one relating models cascading in django?

Background:
I have the below models defined in Django(1.8.5):
class PublishInfo(models.Model):
pass
class Book(models.Model):
info = models.OneToOneField(
PublishInfo, on_delete=models.CASCADE)
class Newspaper(models.Model):
info = models.OneToOneField(
PublishInfo, on_delete=models.CASCADE)
Where Book and NewsPaper shares a same model PublishInfo as a OneToOneField, which is in fact a unique foreign key.
Now, if I delete a PublishInfo Object, the relating Book or Newspaper object is deleted with cascading.
Question:
But in fact, I want to delete the PublishInfo object cascading when I delete the Book or Newspaper object. This way is the way I may call.
Is there any good way to automatically cascading the deletion in the reverse direction in this case? And, if yes, could it be explained?

You attach post_delete signal to your model so it is called upon deletion of an instance of Book or Newspaper:
from django.db.models.signals import post_delete
from django.dispatch import receiver
#receiver(post_delete, sender=Book)
def auto_delete_publish_info_with_book(sender, instance, **kwargs):
instance.info.delete()
#receiver(post_delete, sender=Newspaper)
def auto_delete_publish_info_with_newpaper(sender, instance, **kwargs):
instance.info.delete()

Another straight forward solution by overriding save and delete method:
Comparing to the answer of #ozgur, I found using signal to cascading the delete action has the same effect as deleting by overriding the Model.delete() method, and also we might auto create the attached PublishInfo:
class Book(models.Model):
info = models.OneToOneField(
PublishInfo, on_delete=models.CASCADE)
def save(self, *args, **kwargs):
super().save(*args, **kwargs)
if not self.info:
self.info = Publish.objects.create()
super().save(*args, **kwargs)
def delete(self, *args, **kwargs):
super().delete(*args, **kwargs)
if self.info:
self.info.delete()
More structured and reusable solution:
So, soon I realized the three listing field and methods are obviously redundant on each Model which was attaching the PublishInfo models as a field.
So, why don't we use inheritance?
class PublishInfoAttachedModel(models.Model):
info = models.OneToOneField(
PublishInfo, related_name='$(class)s',
on_delete=models.CASCADE)
def save(self, *args, **kwargs):
super().save(*args, **kwargs)
if not self.info:
self.info = Publish.objects.create()
super().save(*args, **kwargs)
def delete(self, *args, **kwargs):
super().delete(*args, **kwargs)
if self.info:
self.info.delete()
class Meta:
abstract = True
Remember to add abstract = True in its meta class.
So, now we are free to add PublishInfo in any other models we want to attach that model, and we can make more than one such abstract models:
class Book(PublishInfoAttachedModel,
models.Model):
pass
class NewsPaper(PublishInfoAttachedModel,
CommentsAttachedModel, # if we have other attached model info
models.Model):
pass
Notice the models.Model class in the trailing super class list can be ignored, I wrote this is just to make the classes more obvious as a Model.

How to validate uniqueness constraint across foreign key (django)

I have the following (simplified) data structure:
Site
-> Zone
-> Room
-> name
I want the name of each Room to be unique for each Site.
I know that if I just wanted uniqueness for each Zone, I could do:
class Room(models.Model):
zone = models.ForeignKey(Zone)
name = models.CharField(max_length=255)
class Meta:
unique_together = ('name', 'zone')
But I can't do what I really want, which is:
class Room(models.Model):
zone = models.ForeignKey(Zone)
name = models.CharField(max_length=255)
class Meta:
unique_together = ('name', 'zone__site')
I tried adding a validate_unique method, as suggested by this question:
class Room(models.Model):
zone = models.ForeignKey(Zone)
name = models.CharField(max_length=255)
def validate_unique(self, exclude=None):
qs = Room.objects.filter(name=self.name)
if qs.filter(zone__site=self.zone__site).exists():
raise ValidationError('Name must be unique per site')
models.Model.validate_unique(self, exclude=exclude)
but I must be misunderstanding the point/implementation of validate_unique, because it is not being called when I save a Room object.
What would be the correct way to implement this check?

Methods are not called on their own when saving the model.
One way to do this is to have a custom save method that calls the validate_unique method when a model is saved:
class Room(models.Model):
zone = models.ForeignKey(Zone)
name = models.CharField(max_length=255)
def validate_unique(self, exclude=None):
qs = Room.objects.filter(name=self.name)
if qs.filter(zone__site=self.zone__site).exists():
raise ValidationError('Name must be unique per site')
def save(self, *args, **kwargs):
self.validate_unique()
super(Room, self).save(*args, **kwargs)

class Room(models.Model):
zone = models.ForeignKey(Zone)
name = models.CharField(max_length=255)
def validate_unique(self, *args, **kwargs):
super(Room, self).validate_unique(*args, **kwargs)
qs = Room.objects.filter(name=self.name)
if qs.filter(zone__site=self.zone__site).exists():
raise ValidationError({'name':['Name must be unique per site',]})
I needed to make similar program. It worked.

The Django Validation objects documentation explains the steps involved in validation including this snippet
Note that full_clean() will not be called automatically when you call your model's save() method
If the model instance is being created as a result of using a ModelForm, then validation will occur when the form is validated.
There are a some options in how you handle validation.
Call the model instance's full_clean() manually before saving.
Override the save() method of the model to perform validation on every save. You can choose how much validation should occur here, whether you want full validation or only uniqueness checks.
class Room(models.Model):
def save(self, *args, **kwargs):
self.full_clean()
super(Room, self).save(*args, **kwargs)
Use a Django pre_save signal handler which will automatically perform validation before a save. This provides a very simple way to add validation on exisiting models without any additional model code.
# In your models.py
from django.db.models.signals import pre_save
def validate_model_signal_handler(sender, **kwargs):
"""
Signal handler to validate a model before it is saved to database.
"""
# Ignore raw saves.
if not kwargs.get('raw', False):
kwargs['instance'].full_clean()
pre_save.connect(validate_model_signal_handler,
sender=Room,
dispatch_uid='validate_model_room')

How do I restrict foreign keys choices to related objects only in django

I have a two way foreign relation similar to the following
class Parent(models.Model):
name = models.CharField(max_length=255)
favoritechild = models.ForeignKey("Child", blank=True, null=True)
class Child(models.Model):
name = models.CharField(max_length=255)
myparent = models.ForeignKey(Parent)
How do I restrict the choices for Parent.favoritechild to only children whose parent is itself? I tried
class Parent(models.Model):
name = models.CharField(max_length=255)
favoritechild = models.ForeignKey("Child", blank=True, null=True, limit_choices_to = {"myparent": "self"})
but that causes the admin interface to not list any children.

I just came across ForeignKey.limit_choices_to in the Django docs.
Not sure yet how it works, but it might be the right thing here.
Update: ForeignKey.limit_choices_to allows one to specify either a constant, a callable or a Q object to restrict the allowable choices for the key. A constant obviously is of no use here, since it knows nothing about the objects involved.
Using a callable (function or class method or any callable object) seems more promising. However, the problem of how to access the necessary information from the HttpRequest object remains. Using thread local storage may be a solution.
2. Update: Here is what has worked for me:
I created a middleware as described in the link above. It extracts one or more arguments from the request's GET part, such as "product=1", and stores this information in the thread locals.
Next there is a class method in the model that reads the thread local variable and returns a list of ids to limit the choice of a foreign key field.
#classmethod
def _product_list(cls):
"""
return a list containing the one product_id contained in the request URL,
or a query containing all valid product_ids if not id present in URL
used to limit the choice of foreign key object to those related to the current product
"""
id = threadlocals.get_current_product()
if id is not None:
return [id]
else:
return Product.objects.all().values('pk').query
It is important to return a query containing all possible ids if none was selected so that the normal admin pages work ok.
The foreign key field is then declared as:
product = models.ForeignKey(
Product,
limit_choices_to={
id__in=BaseModel._product_list,
},
)
The catch is that you have to provide the information to restrict the choices via the request. I don't see a way to access "self" here.

The 'right' way to do it is to use a custom form. From there, you can access self.instance, which is the current object. Example --
from django import forms
from django.contrib import admin
from models import *
class SupplierAdminForm(forms.ModelForm):
class Meta:
model = Supplier
fields = "__all__" # for Django 1.8+
def __init__(self, *args, **kwargs):
super(SupplierAdminForm, self).__init__(*args, **kwargs)
if self.instance:
self.fields['cat'].queryset = Cat.objects.filter(supplier=self.instance)
class SupplierAdmin(admin.ModelAdmin):
form = SupplierAdminForm

The new "right" way of doing this, at least since Django 1.1 is by overriding the AdminModel.formfield_for_foreignkey(self, db_field, request, **kwargs).
See http://docs.djangoproject.com/en/1.2/ref/contrib/admin/#django.contrib.admin.ModelAdmin.formfield_for_foreignkey
For those who don't want to follow the link below is an example function that is close for the above questions models.
class MyModelAdmin(admin.ModelAdmin):
def formfield_for_foreignkey(self, db_field, request, **kwargs):
if db_field.name == "favoritechild":
kwargs["queryset"] = Child.objects.filter(myparent=request.object_id)
return super(MyModelAdmin, self).formfield_for_manytomany(db_field, request, **kwargs)
I'm only not sure about how to get the current object that is being edited. I expect it is actually on the self somewhere but I'm not sure.

This isn't how django works. You would only create the relation going one way.
class Parent(models.Model):
name = models.CharField(max_length=255)
class Child(models.Model):
name = models.CharField(max_length=255)
myparent = models.ForeignKey(Parent)
And if you were trying to access the children from the parent you would do
parent_object.child_set.all(). If you set a related_name in the myparent field, then that is what you would refer to it as. Ex: related_name='children', then you would do parent_object.children.all()
Read the docs http://docs.djangoproject.com/en/dev/topics/db/models/#many-to-one-relationships for more.

If you only need the limitations in the Django admin interface, this might work. I based it on this answer from another forum - although it's for ManyToMany relationships, you should be able to replace formfield_for_foreignkey for it to work. In admin.py:
class ParentAdmin(admin.ModelAdmin):
def get_form(self, request, obj=None, **kwargs):
self.instance = obj
return super(ParentAdmin, self).get_form(request, obj=obj, **kwargs)
def formfield_for_foreignkey(self, db_field, request=None, **kwargs):
if db_field.name == 'favoritechild' and self.instance:
kwargs['queryset'] = Child.objects.filter(myparent=self.instance.pk)
return super(ChildAdmin, self).formfield_for_foreignkey(db_field, request=request, **kwargs)

#Ber: I have added validation to the model similar to this
class Parent(models.Model):
name = models.CharField(max_length=255)
favoritechild = models.ForeignKey("Child", blank=True, null=True)
def save(self, force_insert=False, force_update=False):
if self.favoritechild is not None and self.favoritechild.myparent.id != self.id:
raise Exception("You must select one of your own children as your favorite")
super(Parent, self).save(force_insert, force_update)
which works exactly how I want, but it would be really nice if this validation could restrict choices in the dropdown in the admin interface rather than validating after the choice.

I'm trying to do something similar. It seems like everyone saying 'you should only have a foreign key one way' has maybe misunderstood what you're trying do.
It's a shame the limit_choices_to={"myparent": "self"} you wanted to do doesn't work... that would have been clean and simple. Unfortunately the 'self' doesn't get evaluated and goes through as a plain string.
I thought maybe I could do:
class MyModel(models.Model):
def _get_self_pk(self):
return self.pk
favourite = models.ForeignKey(limit_choices_to={'myparent__pk':_get_self_pk})
But alas that gives an error because the function doesn't get passed a self arg :(
It seems like the only way is to put the logic into all the forms that use this model (ie pass a queryset in to the choices for your formfield). Which is easily done, but it'd be more DRY to have this at the model level. Your overriding the save method of the model seems a good way to prevent invalid choices getting through.
Update
See my later answer for another way https://stackoverflow.com/a/3753916/202168

Do you want to restrict the choices available in the admin interface when creating/editing a model instance?
One way to do this is validation of the model. This lets you raise an error in the admin interface if the foreign field is not the right choice.
Of course, Eric's answer is correct: You only really need one foreign key, from child to parent here.

An alternative approach would be not to have 'favouritechild' fk as a field on the Parent model.
Instead you could have an is_favourite boolean field on the Child.
This may help:
https://github.com/anentropic/django-exclusivebooleanfield
That way you'd sidestep the whole problem of ensuring Children could only be made the favourite of the Parent they belong to.
The view code would be slightly different but the filtering logic would be straightforward.
In the admin you could even have an inline for Child models that exposed the is_favourite checkbox (if you only have a few children per parent) otherwise the admin would have to be done from the Child's side.

A much simpler variation of #s29's answer:
Instead of customising the form,
You can simply restrict the choices available in form field from your view:
what worked for me was:
in forms.py:
class AddIncomingPaymentForm(forms.ModelForm):
class Meta:
model = IncomingPayment
fields = ('description', 'amount', 'income_source', 'income_category', 'bank_account')
in views.py:
def addIncomingPayment(request):
form = AddIncomingPaymentForm()
form.fields['bank_account'].queryset = BankAccount.objects.filter(profile=request.user.profile)

from django.contrib import admin
from sopin.menus.models import Restaurant, DishType
class ObjInline(admin.TabularInline):
def __init__(self, parent_model, admin_site, obj=None):
self.obj = obj
super(ObjInline, self).__init__(parent_model, admin_site)
class ObjAdmin(admin.ModelAdmin):
def get_inline_instances(self, request, obj=None):
inline_instances = []
for inline_class in self.inlines:
inline = inline_class(self.model, self.admin_site, obj)
if request:
if not (inline.has_add_permission(request) or
inline.has_change_permission(request, obj) or
inline.has_delete_permission(request, obj)):
continue
if not inline.has_add_permission(request):
inline.max_num = 0
inline_instances.append(inline)
return inline_instances
class DishTypeInline(ObjInline):
model = DishType
def formfield_for_foreignkey(self, db_field, request=None, **kwargs):
field = super(DishTypeInline, self).formfield_for_foreignkey(db_field, request, **kwargs)
if db_field.name == 'dishtype':
if self.obj is not None:
field.queryset = field.queryset.filter(restaurant__exact = self.obj)
else:
field.queryset = field.queryset.none()
return field
class RestaurantAdmin(ObjAdmin):
inlines = [
DishTypeInline
]