Unregistered Pydantic Type Error

Description

This error is thrown during schema construction when a field uses a Pydantic model that is neither a Strawberry type nor registered in the schema’s scalar_map . It applies to regular Strawberry fields as well as fields of strawberry.pydantic types (with Pydantic 2.11 or newer):

import pydantic
import strawberry
 
 
class Address(pydantic.BaseModel):
    street: str
 
 
@strawberry.type
class Query:
    @strawberry.field
    def address(self) -> Address:
        return Address(street="Main Street")
 
 
schema = strawberry.Schema(query=Query)

This happens because Strawberry needs to know the GraphQL type of every field, and Address hasn’t been converted to a GraphQL type.

How to fix this error

You can fix this error by decorating the model used by the field, with @strawberry.pydantic.type for output types and @strawberry.pydantic.input for inputs:

import pydantic
import strawberry
 
 
@strawberry.pydantic.type
class Address(pydantic.BaseModel):
    street: str
 
 
@strawberry.pydantic.type
class User(pydantic.BaseModel):
    name: str
    address: Address

Subclasses of decorated models need their own decorator too, as they can have fields of their own. Otherwise the field would silently use the base model's GraphQL type, without the subclass's fields:

@strawberry.pydantic.type
class Animal(pydantic.BaseModel):
    name: str
 
 
@strawberry.pydantic.type
class Dog(Animal):
    breed: str
 
 
@strawberry.pydantic.type
class Owner(pydantic.BaseModel):
    dog: Dog

A parametrized generic model, like Page[User] , is a GraphQL type when its generic model is decorated, so decorate the generic model:

from typing import Generic, TypeVar
 
T = TypeVar("T")
 
 
@strawberry.pydantic.type
class Page(pydantic.BaseModel, Generic[T]):
    items: list[T]
 
 
@strawberry.pydantic.type
class Directory(pydantic.BaseModel):
    users: Page[User]

If the model has a type from the experimental integration, strawberry.experimental.pydantic , the error says it only has a strawberry.experimental.pydantic type: the two integrations can’t use each other’s types. Decorate the model with @strawberry.pydantic.type too, with a different name if its experimental type is also in the schema:

import pydantic
import strawberry
 
 
class AddressModel(pydantic.BaseModel):
    street: str
 
 
@strawberry.experimental.pydantic.type(model=AddressModel, all_fields=True)
class Address:
    pass
 
 
strawberry.pydantic.type(AddressModel, name="NewAddress")
 
 
@strawberry.pydantic.type
class User(pydantic.BaseModel):
    name: str
    address: AddressModel