Django Admin: OneToOne Relation as an Inline?

Question

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',)

Asked By: Jiaaro

||

Source

Answer 1

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

Answered By: Daniel Roseman

Answer 2

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.

Answered By: Henri

Answer 3

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.

Answered By: Alexey

Answer 4

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

Answered By: stephendwolff

Answer 5

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, )

Answered By: Kai – Kazuya Ito

Django Admin: OneToOne Relation as an Inline?

Question:

Answers:

Jun, 2022 Update: