python - Pydantic 设置属性与默认函数

Question

是否可以为不可变 Pydantic Models 传递函数 setter ？ .

例如:

from uuid import uuid4, UUID
from pydantic import BaseModel
from datetime import datetime

def generate_uuid():
    return uuid4()

def get_datetimenow():
    return datetime.now()

class Item(BaseModel):

    class Config:
        allow_mutation = False
        extra = "forbid"

    id: UUID
    created_at: datetime

我想要方法 generate_uuid 和 get_datetimenow 来设置属性。

>>> Item()
ValidationError: 2 validation errors for Item
id
  field required (type=value_error.missing)
created_at
  field required (type=value_error.missing)

虽然我想获得一个 Item 对象，并自动设置了 id 和 created_at。与运行时相同的结果:

>>> Item(id=generate_uuid(), created_at=get_datetimenow())
Item(id=UUID('8f898730-3fad-4ca9-9667-c090f62a2954'), created_at=datetime.datetime(2021, 1, 19, 21, 13, 7, 58051))

Answer 1

您可以使用 default_factory Field 的参数与任意函数。像这样:

from uuid import uuid4, UUID
from pydantic import BaseModel, Field
from datetime import datetime

class Item(BaseModel):

    class Config:
        allow_mutation = False
        extra = "forbid"

    id: UUID = Field(default_factory=uuid4)
    created_at: datetime = Field(default_factory=datetime.now)