从**kwargs在构造期间设置类的属性

这里是python noob,

目前我正在使用sqlacalchemy,我有:

from __init__ import Base
from sqlalchemy.schema import Column, ForeignKey
from sqlalchemy.types import Integer, String
from sqlalchemy.orm import relationship

class User(Base):
    __tablename__ = "users"
    id = Column(Integer, primary_key=True)
    username = Column(String, unique=True)
    email = Column(String)
    password = Column(String)
    salt = Column(String)
    openids = relationship("OpenID", backref="users")

User.__table__.create(checkfirst=True)

#snip definition of OpenID class

def create(**kwargs):
    user = User()
    if "username" in kwargs.keys():
        user.username = kwargs['username']
    if "email" in kwargs.keys():
        user.username = kwargs['email']
    if "password" in kwargs.keys():
        user.password = kwargs['password']

    return user

这是在 /db/users.py ,因此其用途如下:

from db import users
new_user = users.create(username="Carson", password="1234")
new_user.email = "email@address.com"
users.add(new_user) #this function obviously not defined yet

但是代码在 create() 有点蠢,我想知道是否有更好的方法可以做到这一点,不需要一个if梯形图,如果添加了用户对象中没有的任何键,这将失败。像:

for attribute in kwargs.keys():
    if attribute in User:
        setattr(user, attribute, kwargs[attribute])
    else:
        raise Exception("blah")

这样我就可以把它放到它自己的函数中(除非希望已经存在了一个函数?)因此,我不必一次又一次地执行if梯形图,这样就可以在不修改此代码的情况下更改表结构。

有什么建议吗?