我正在构建一个存储具有一组属性的电子邮件的模型:
email_list_table = Table('email_list', Base.metadata,
Column('email_id', String, ForeignKey('emails.id')),
Column('list_id', Integer, ForeignKey('lists.id'))
)
class Email(Base):
__tablename__ = 'emails'
id = Column(String, primary_key=True)
from_email = Column(String, nullable=False)
from_name = Column(String, nullable=False)
date = Column(DateTime, nullable=False)
in_reply_to = Column(String)
subject = Column(String, nullable=False)
content = Column(String, nullable=False)
lists = relationship("List", secondary=email_list_table)
references = relationship("Email") # This is the issue
def __repr__(self):
return "<Email(id='%s', from='%s', subject='%s')>" % (self.id, self.from_name, self.subject)
class List(Base):
__tablename__ = 'lists'
id = Column(Integer, primary_key=True)
name = Column(String, nullable=False)
email = Column(String, nullable=False)
description = Column(String, nullable=False)
def __repr__(self):
return "<List(id='%s')>" % (self.id)
我想将references
关系表示为电子邮件列表。例如,电子邮件可以引用多个其他电子邮件,并被多个电子邮件引用。
在传统的SQL中,我会使用第二个表,其中包含referrer_id
和referenced_id
的复合主键,在查询时,我会加入这两个表来获取引用的电子邮件列表。这样可以很容易地找到给定ID引用的电子邮件以及给定ID引用的电子邮件。
我查看了SQLAlchemy上Adjacency List Relationships的文档,但我不确定那里的示例是否适用于我的模型(我想我可以为referrer_id
添加一列,但那不会&# 39;使用&#39; join&#39;表生成与我预期相同的表格结构。
使用SQLAlchemy制作模型的最佳/最正确方法是什么?
答案 0 :(得分:0)
我通过创建第二个表并在Email
类中引用它来得到我想要的答案:
email_references = Table('email_references', Base.metadata,
Column('referencer_id', String, ForeignKey('emails.id'), primary_key=True),
Column('referenced_id', String, ForeignKey('emails.id'), primary_key=True)
)
class Email(Base):
__tablename__ = 'emails'
id = Column(String, primary_key=True)
from_email = Column(String, nullable=False)
from_name = Column(String, nullable=False)
date = Column(DateTime, nullable=False)
in_reply_to = Column(String)
subject = Column(String, nullable=False)
content = Column(String, nullable=False)
lists = relationship("List", secondary=email_list_table)
references = relationship("Email",
secondary=email_references,
primaryjoin=id==email_references.c.referencer_id,
secondaryjoin=id==email_references.c.referenced_id
)