如果没有传递，如何将默认值传递给变量？
如果没有传递，如何将默认值传递给变量？
提问于 2020-08-27 13:12:46
如果字段中没有传递，我可以在Pydantic中创建默认值吗？
我有以下代码，但在我看来，这里的验证器只用于模型的初始化，而不是其他的。
我的代码：
class User(BaseModel):
name: Optional[str] = ''
password: Optional[str] = ''
email: EmailStr
@validator('name')
def set_name(cls, name):
return name or 'foo'遇到的问题：
user = User(name=None, password='some_password', email='user@example.com')
print("Name is ", user.name)
# > 'Name is foo'
user.name = None
print("Name is ", user.name)
# > 'Name is None'期望输出：
user = User(name='some_name', password='some_password', email='user@example.com')
user.name = None
print("Name is ", user.name)
# > 'Name is foo'关于如何获得期望的输出，有什么想法吗？我认为拥有getter和setter将有助于解决这个问题。然而，我无法让他们在毕丹式的模型中工作：
尝试实现getter和setter的：
class User(BaseModel):
name: Optional[str] = ''
password: Optional[str] = ''
email: EmailStr
def get_password(self):
return self.password
def set_password(self, password):
self.password = hash_password(password)
password = property(get_password, set_password)
user = User(name='some_name', password='some_password', email='user@example.com')
# > RecursionError: maximum recursion depth exceeded我也试过房地产装饰师：
class User(BaseModel):
name: Optional[str] = ''
password: Optional[str] = ''
email: EmailStr
@property
def password(self):
return self._password
@password.setter
def password(self, password):
pwd_context = CryptContext(schemes=["bcrypt"], deprecated="auto")
self._password = pwd_context.hash(password)
user = User(name='some_name', email='user@example.com')
user.password = 'some_password'
# > ValueError: "User" object has no field "password"我也尝试过覆盖init:
class User(BaseModel):
name: Optional[str] = ""
password: Optional[str] = ""
email: EmailStr
def __init__(self, name, password, email):
pwd_context = CryptContext(schemes=["bcrypt"], deprecated="auto")
password = pwd_context.hash(password)
super().__init__(name=name, password=password, email=email)
user = User(name="some_name", password="some_password", email='user@example.com')
print(user.password)
# > AYylwSnbQgCHrl4uue6kO7yiuT20lazSzK7x # Works as expected
user.password = "some_other_password"
print(user.password)
# > "some_other_password" # Does not work
user.password = None
print(user.password)
# > None # Does not work eitherStack Overflow用户
回答已采纳
发布于 2020-08-27 13:33:01
您需要在模型配置中启用validate_assignment选项：
from typing import Optional
from pydantic import BaseModel, validator
class User(BaseModel):
name: Optional[str] = ''
password: Optional[str] = ''
class Config:
validate_assignment = True
@validator('name')
def set_name(cls, name):
return name or 'foo'
user = User(name=None, password='some_password', )
print("Name is ", user.name)
user.name = None
print("Name is ", user.name)Name is foo
Name is foo页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/63616798
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