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我试图创建一个人,一个公司和一个人在给定公司使用的电子邮件地址之间的链接。SQLAlchemy关于外键和关联表的问题
在SQL术语中,我认为这将是四个表,Person,Company,Email和email_company_association,其中两个FK在email_company_association中,一个在电子邮件中,一个在公司。然后将电子邮件中的一个FK发送给Person。电子邮件地址也可以直接连接到个人,因此电子邮件中的个人FK也是如此。
但是我有点茫然如何在SQLAlchemy中做到这一点。
我已经试过类似:
class Person(Base):
"""
A person
"""
__tablename__ = 'persons'
id = Column(Integer, Sequence('person_id_seq'), primary_key=True)
name = Column(String(255), unique=True)
surename = Column(String(255))
forename = Column(String(255))
class Company(Base):
"""
A company
"""
__tablename__ = 'companies'
id = Column(Integer, Sequence('company_id_seq'), primary_key=True)
name = Column(String)
email_addresses = relationship("Company_Email_Association", backref="company")
person_id = Column(Integer, ForeignKey('persons.id'), nullable=False)
person = relationship("Person", backref=backref('companies', order_by=id))
class Email(Base):
"""
Email address
"""
__tablename__ = 'emailaddresses'
id = Column(Integer, primary_key=True)
email = Column(String, nullable=False)
person_id = Column(Integer, ForeignKey('persons.id'), nullable=False)
person = relationship("Person", backref=backref('emailaddresses', order_by=id))
class Company_Email_Association(Base):
__tablename__ = 'company_email_assoc'
company_id = Column(Integer, ForeignKey('companies.id'), primary_key=True)
email_id = Column(Integer, ForeignKey('emailaddresses.id'), primary_key=True)
email = relationship("Email")
我使用此约像这样:
p = Person()
c = Company(name="Foo LTD")
cea = Company_Email_Association()
cea.email = Email(email="[email protected]") # This breaks since Email needs persons.id
c.email_addresses.append(cea)
p.companies.append(c)
这是我的错误:
sqlalchemy.exc.IntegrityError: (IntegrityError) null value in column "person_id" violates not-null constraint
DETAIL: Failing row contains (1, [email protected], null).
'INSERT INTO emailaddresses (email, person_id) VALUES (%(email)s, %(person_id)s) RETURNING emailaddresses.id' {'person_id': None,'email': '[email protected]'}
我猜我做两个SQLAlchemy的和造型非常错误的,但我在在做什么损失。