如何在FastAPI正文验证中使用区分的联合类型？（模型联盟）

Question

我从Typescript中知道一个概念，叫做有区别的联合。这是一个放置2个结构（类等）的地方，类型取决于结构的值。我正试图通过Pydantic验证在FastAPI中实现类似的功能。我可以接收两种不同的请求有效负载。它是一个还是另一个取决于accountType变量。如果是creative，则应通过RegistrationPayloadCreative进行验证，如果是brand，则应通过RegistrationPayloadBrand进行验证。我如何做到这一点？找不到其他解决方案

问题是它要么返回

unexpected value; permitted: 'creative' (type=value_error.const; given=brand; permitted=('creative',))

或者根本不起作用

class RegistrationPayloadBase(BaseModel):
    first_name: str
    last_name: str
    email: str
    password: str


class RegistrationPayloadCreative(RegistrationPayloadBase):
    accountType: Literal['creative']


class RegistrationPayloadBrand(RegistrationPayloadBase):
    company: str
    phone: str
    vat: str
    accountType: Literal['brand']

class A(BaseModel):
    b: Union[RegistrationPayloadBrand, RegistrationPayloadCreative]

def main():
    A(b={'first_name': 'sdf', 'last_name': 'sdf', 'email': 'sdf', 'password': 'sdfds', 'accountType': 'brand'})

if __name__ == '__main__':
    main()

Answer 1

该错误消息有点误导，因为问题在于公司/电话/增值税字段在RegistrationPayloadBrand中是强制性的

因此：

 >>> A(b={'first_name': 'sdf', 'last_name': 'sdf', 'email': 'sdf', 'password': 'sdfds', 'accountType': 'brand', 'company':'foo','vat':'bar', 'phone':'baz'}) 
A(b=RegistrationPayloadBrand(first_name='sdf', last_name='sdf',
email='sdf', password='sdfds', company='foo', phone='baz', vat='bar', accountType='brand'))

>>> A(b={'first_name': 'sdf', 'last_name': 'sdf', 'email': 'sdf', 'password': 'sdfds', 'accountType': 'creative'})
A(b=RegistrationPayloadCreative(first_name='sdf', last_name='sdf', email='sdf', password='sdfds', accountType='creative'))

或者使其可选（如果有效负载不一定包含这些字段）

class RegistrationPayloadBrand(RegistrationPayloadBase):
    company: Optional[str]
    phone:   Optional[str]
    vat:     Optional[str]
    accountType: Literal['brand']

>>> A(b={'first_name': 'sdf', 'last_name': 'sdf', 'email': 'sdf', 'password': 'sdfds', 'accountType': 'brand'})
A(b=RegistrationPayloadBrand(first_name='sdf', last_name='sdf', email='sdf', password='sdfds', company=None, phone=None, vat=None, accountType='brand'))

>>> A(b={'first_name': 'sdf', 'last_name': 'sdf', 'email': 'sdf', 'password': 'sdfds', 'accountType': 'creative'})
A(b=RegistrationPayloadCreative(first_name='sdf', last_name='sdf', email='sdf', password='sdfds', accountType='creative'))

这将解决问题

Answer 2

您应该使用__root__和parse_obj

from typing import Union

from pydantic import BaseModel


class PlanetItem(BaseModel):
    id: str
    planet_name: str 
    # ...

class CarItem(BaseModel):
    id: str
    name: str
    # ...

class EitherItem(BaseModel):
    __root__: Union[PlanetItem, CarItem]



@app.get("/items/{item_id}", response_model=EitherItem)
def get_items(item_id):
    return EitherItem.parse_obj(response) # Now you get either PlanetItem or CarItem

信用证：https://github.com/tiangolo/fastapi/issues/2279#issuecomment-787517707