即使一个失败，如何通过所有 Pydantic 验证器，然后在 FastAPI 响应中引发多个 ValueErrors？

Question

是否可以调用所有验证器以返回完整的错误列表？

@validator('password', always=True)
def validate_password1(cls, value):
    password = value.get_secret_value()

    min_length = 8
    if len(password) < min_length:
        raise ValueError('Password must be at least 8 characters long.')

    return value

@validator('password', always=True)
def validate_password2(cls, value):
    password = value.get_secret_value()

    if not any(character.islower() for character in password):
        raise ValueError('Password should contain at least one lowercase character.')

    return value

当前行为似乎一次调用一个验证器。

我的 Pydantic class:

class User(BaseModel):
    email: EmailStr
    password: SecretStr

如果我没有在请求中包含 email 或 password 字段，那么我会在数组中得到两个验证失败，这就是我想为 password 字段，但当前行为似乎调用一个，如果失败则立即抛出错误。

Answer 1

你不能那样使用 Pydantic's validators；它看起来总是其中之一。

为了得到你的答案，你可以使用以下两种方法

1 - 您可以使用一个检查所有条件的主验证器

@validator('password', always=True)
def validate_password(cls, value):
    password = value.get_secret_value()

    validate_password1(password)
    validate_password2(password)

    return value
    
def validate_password1(password):

    min_length = 8
    if len(password) < min_length:
        raise ValueError('Password must be at least 8 characters long.')


def validate_password2(password):

    if not any(character.islower() for character in password):
        raise ValueError('Password should contain at least one lowercase character.')

2 - 您可以在模型中使用重复变量来检查条件

class User(BaseModel):
    email: EmailStr
    password: SecretStr
    password2: SecretStr

显然，您的装饰器应该是：

@validator('password', always=True)
def validate_password1(cls, value):

和

@validator('password2', always=True)
def validate_password2(cls, value):

更新： OP 想提出所有错误，所以更新后的答案如下。

除了第一个项目符号，您还可以尝试类似的方法：

@validator('password', always=True)
def validate_password(cls, value):
    password = value.get_secret_value()

    try:
        validate_password1(password)
    except Exception as e:
        print('First error: ' + str(e))
        
    try:
        validate_password2(password)
    except Exception as e:
        print('Second error: ' + str(e))

    return value
    
def validate_password1(password):

    min_length = 8
    if len(password) < min_length:
        raise ValueError('Password must be at least 8 characters long.')


def validate_password2(password):

    if not any(character.islower() for character in password):
        raise ValueError('Password should contain at least one lowercase character.')

但是，返回 value 时要小心。您可以尝试在主代码中再添加一个自定义异常。

Answer 2

您不能以您在问题中展示的方式为特定领域提出多个 Validation errors/exceptions。下面给出了建议的解决方案。

选项 1

使用单个变量连接错误消息，并在末尾引发一次 ValueError（如果发生错误）：

@validator('password', always=True)
def validate_password1(cls, value):
    password = value.get_secret_value()
    min_length = 8
    errors = ''
    if len(password) < min_length:
        errors += 'Password must be at least 8 characters long. '
    if not any(character.islower() for character in password):
        errors += 'Password should contain at least one lowercase character.'
    if errors:
        raise ValueError(errors)
        
    return value

在满足上述所有条件语句的情况下，输出将是：

{
  "detail": [
    {
      "loc": [
        "body",
        "password"
      ],
      "msg": "Password must be at least 8 characters long. Password should contain at least one lowercase character.",
      "type": "value_error"
    }
  ]
}

选项 2

提高 ValidationError directly, using a list of ErrorWrapper class.

from pydantic import ValidationError
from pydantic.error_wrappers import ErrorWrapper

@validator('password', always=True)
def validate_password1(cls, value):
    password = value.get_secret_value()
    min_length = 8
    errors = []
    if len(password) < min_length:
        errors.append(ErrorWrapper(ValueError('Password must be at least 8 characters long.'), loc=None))
    if not any(character.islower() for character in password):
        errors.append(ErrorWrapper(ValueError('Password should contain at least one lowercase character.'), loc=None))
    if errors:
        raise ValidationError(errors, model=User)
        
    return value

由于 FastAPI 似乎正在添加 loc 属性本身，因此 loc 最终将具有 field 名称（即 password）两次，如果它是使用 loc 属性（这是必需参数）添加到 ErrorWrapper 中。因此，您可以将其留空（使用 None），稍后您可以通过 validation exception handler 将其删除，如下所示：

from fastapi import Request, status
from fastapi.encoders import jsonable_encoder
from fastapi.exceptions import RequestValidationError
from fastapi.responses import JSONResponse

@app.exception_handler(RequestValidationError)
async def validation_exception_handler(request: Request, exc: RequestValidationError):
    for error in exc.errors(): 
        error['loc'] = [x for x in error['loc'] if x]  # remove null attributes
        
    return JSONResponse(status_code=status.HTTP_422_UNPROCESSABLE_ENTITY, content=jsonable_encoder({"detail": exc.errors()}))

在满足上述所有条件语句的情况下，输出将是：

{
  "detail": [
    {
      "loc": [
        "body",
        "password"
      ],
      "msg": "Password must be at least 8 characters long.",
      "type": "value_error"
    },
    {
      "loc": [
        "body",
        "password"
      ],
      "msg": "Password should contain at least one lowercase character.",
      "type": "value_error"
    }
  ]
}

即使一个失败，如何通过所有 Pydantic 验证器，然后在 FastAPI 响应中引发多个 ValueErrors？

How to go through all Pydantic validators even if one fails, and then raise multiple ValueErrors in a FastAPI response?

python

python-3.x

pydantic

fastapi

选项 1

选项 2