如果 pydantic 模型定义了别名，如何使用 `from_orm`？

Question

虽然 pydantic 的 ORM 模式已记录 here，但不幸的是，没有关于使用别名的文档。

如果 pydantic 模型定义了别名，如何使用 from_orm？

如果别名存在，from_orm 工厂似乎忘记了所有非别名。 - 请参阅下面的错误消息和相应的代码。这是错误还是功能？

下面的代码片段意外失败并出现验证错误：

pydantic.error_wrappers.ValidationError: 1 validation error for SimpleModel
threeWordsId
field required (type=value_error.missing)

from sqlalchemy import Column, String
from sqlalchemy.ext.declarative import declarative_base
from pydantic import BaseModel, Field

Base = declarative_base()

class SimpleOrm(Base):
    __tablename__ = 'simples'
    three_words_id = Column(String, primary_key=True)

class SimpleModel(BaseModel):
    three_words_id: str = Field(..., alias="threeWordsId")

    class Config:
        orm_mode=True

simple_orm = SimpleOrm(three_words_id='abc')
simple_oops = SimpleModel.from_orm(simple_orm)

Answer 1

在配置中使用 allow_population_by_field_name = True。

喜欢

from sqlalchemy import Column, String
from sqlalchemy.ext.declarative import declarative_base
from pydantic import BaseModel, Field

Base = declarative_base()


class SimpleOrm(Base):
    __tablename__ = 'simples'
    three_words_id = Column(String, primary_key=True)


class SimpleModel(BaseModel):
    three_words_id: str = Field(..., alias="threeWordsId")

    class Config:
        orm_mode = True
        allow_population_by_field_name = True
        # allow_population_by_alias = True # in case pydantic.version.VERSION < 1.0


simple_orm = SimpleOrm(three_words_id='abc')
simple_oops = SimpleModel.from_orm(simple_orm)

print(simple_oops.json())  # {"three_words_id": "abc"}
print(simple_oops.json(by_alias=True))  # {"threeWordsId": "abc"}


from fastapi import FastAPI

app = FastAPI()


@app.get("/model", response_model=SimpleModel)
def get_model():
    # results in {"threeWordsId":"abc"}
    return SimpleOrm(three_words_id='abc')

如果 pydantic 模型定义了别名，如何使用 `from_orm`？

How to use `from_orm` if the pydantic model defines aliases?

orm

sqlalchemy

pydantic