构造函数和 Pydantic

Question

我想创建一个带有构造函数的 Pydantic 类，该构造函数对输入进行一些数学运算并相应地设置对象变量：

class PleaseCoorperate(BaseModel):
    self0: str
    next0: str

    def __init__(self, page: int, total: int, size: int):
        # Do some math here and later set the values
        self.self0 = ""
        self.next0 = ""
    

但是当我尝试实际使用该类时，page = PleaseCoorperate(0, 1, 50)我收到此错误：

main_test.py:16: 
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 
main.py:46: in __init__
    self.self0 = ""
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 

>   ???
E   AttributeError: __fields_set__

这里发生了什么？我可以不在Pydantic类上使用构造函数吗？

Answer 1

您没有调用BaseModel构造函数，因此，您跳过了将类变量转换为属性的所有 pydantic 魔法。这：

class PleaseCoorperate(BaseModel):
    self0: str
    next0: str

    def __init__(self, page: int, total: int, size: int):
        # Do some math here and later set the values
        self0 = ""
        next0 = ""
        super().__init__(self0=self0, next0=next0)

应该可以解决问题。但是，我可以建议一个更好的选择，它不会使用类方法覆盖 pydantic 中非常优秀的默认构造函数：

class PleaseCoorperate(BaseModel):
    self0: str
    next0: str

    @classmethod
    def custom_init(cls, page: int, total: int, size: int):
        # Do some math here and later set the values
        self0 = ""
        next0 = ""
        return cls(self0=self0, next0=next0)

x = PleaseCoorperate.custom_init(10, 10, 10)

Answer 2

您可以使用 pydantic 验证器。这些是您解决方案的完美候选者。

from pydantic import BaseModel, validator

class PleaseCoorperate(BaseModel):
    self0: str
    next0: str

    @validator('self0')
    def self0_math_test(cls, v):  # v set the values passed for self0
        # your math here
        return new_self0

    @validator('next0', always=True)  # always if you want to run it even when next0 is not passed (optional)
    def next0_must_have(cls, v, values, **kwargs):  # values sets other variable values
        # your math here
        return new_next0

验证器文档：此处