[英]SQLAlchemy relationship on many-to-many association table
I am trying to build a relationship to another many-to-many relationship, the code looks like this: 我正在尝试与另一个多对多关系建立关系,代码如下所示:
from sqlalchemy import Column, Integer, ForeignKey, Table, ForeignKeyConstraint, create_engine
from sqlalchemy.orm import relationship, backref, scoped_session, sessionmaker
from sqlalchemy.ext.declarative import declarative_base
Base = declarative_base()
supervision_association_table = Table('supervision', Base.metadata,
Column('supervisor_id', Integer, ForeignKey('supervisor.id'), primary_key=True),
Column('client_id', Integer, ForeignKey('client.id'), primary_key=True)
)
class User(Base):
__tablename__ = 'user'
id = Column(Integer, primary_key=True)
class Supervisor(User):
__tablename__ = 'supervisor'
__mapper_args__ = {'polymorphic_identity': 'supervisor'}
id = Column(Integer, ForeignKey('user.id'), primary_key = True)
schedules = relationship("Schedule", backref='supervisor')
class Client(User):
__tablename__ = 'client'
__mapper_args__ = {'polymorphic_identity': 'client'}
id = Column(Integer, ForeignKey('user.id'), primary_key = True)
supervisor = relationship("Supervisor", secondary=supervision_association_table,
backref='clients')
schedules = relationship("Schedule", backref="client")
class Schedule(Base):
__tablename__ = 'schedule'
__table_args__ = (
ForeignKeyConstraint(['client_id', 'supervisor_id'], ['supervision.client_id', 'supervision.supervisor_id']),
)
id = Column(Integer, primary_key=True)
client_id = Column(Integer, nullable=False)
supervisor_id = Column(Integer, nullable=False)
engine = create_engine('sqlite:///temp.db')
db_session = scoped_session(sessionmaker(bind=engine))
Base.metadata.create_all(bind=engine)
What I want to do is to relate a schedule to a specific Client-Supervisor-relationship, though I have not found out how to do it. 我想要做的是将时间表与特定的客户 - 主管关系联系起来,尽管我还没有找到如何做到这一点。 Going through the SQLAlchemy documentation I found a few hints, resulting in the ForeignKeyConstraint on the Schedule-Table.
通过SQLAlchemy文档,我发现了一些提示,导致Schedule-Table上的ForeignKeyConstraint。
How can I specify the relationship to have this association work? 如何指定关联以使此关联有效?
You need to map supervision_association_table so that you can create relationships to/from it. 您需要映射supervision_association_table,以便您可以创建与之关系。
I may be glossing over something here, but it seems like since you have many-to-many here you really can't have Client.schedules - if I say Client.schedules.append(some_schedule), which row in "supervision" is it pointing to ? 我可能会在这里讨论一些事情,但似乎因为你在这里有多对多你真的不能有Client.schedules - 如果我说Client.schedules.append(some_schedule),那么“监督”中的哪一行是它指向? So the example below provides a read-only "rollup" accessor for those which joins the Schedule collections of each SupervisorAssociation.
因此,下面的示例为那些加入每个SupervisorAssociation的Schedule集合的访问者提供了只读“汇总”访问器。 The association_proxy extension is used to conceal, when convenient, the details of the SupervisionAssociation object.
association_proxy扩展用于在方便时隐藏SupervisionAssociation对象的详细信息。
from sqlalchemy import Column, Integer, ForeignKey, Table, ForeignKeyConstraint, create_engine
from sqlalchemy.orm import relationship, backref, scoped_session, sessionmaker
from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy.ext.associationproxy import association_proxy
from itertools import chain
Base = declarative_base()
class SupervisionAssociation(Base):
__tablename__ = 'supervision'
supervisor_id = Column(Integer, ForeignKey('supervisor.id'), primary_key=True)
client_id = Column(Integer, ForeignKey('client.id'), primary_key=True)
supervisor = relationship("Supervisor", backref="client_associations")
client = relationship("Client", backref="supervisor_associations")
schedules = relationship("Schedule")
class User(Base):
__tablename__ = 'user'
id = Column(Integer, primary_key=True)
class Supervisor(User):
__tablename__ = 'supervisor'
__mapper_args__ = {'polymorphic_identity': 'supervisor'}
id = Column(Integer, ForeignKey('user.id'), primary_key = True)
clients = association_proxy("client_associations", "client",
creator=lambda c: SupervisionAssociation(client=c))
@property
def schedules(self):
return list(chain(*[c.schedules for c in self.client_associations]))
class Client(User):
__tablename__ = 'client'
__mapper_args__ = {'polymorphic_identity': 'client'}
id = Column(Integer, ForeignKey('user.id'), primary_key = True)
supervisors = association_proxy("supervisor_associations", "supervisor",
creator=lambda s: SupervisionAssociation(supervisor=s))
@property
def schedules(self):
return list(chain(*[s.schedules for s in self.supervisor_associations]))
class Schedule(Base):
__tablename__ = 'schedule'
__table_args__ = (
ForeignKeyConstraint(['client_id', 'supervisor_id'],
['supervision.client_id', 'supervision.supervisor_id']),
)
id = Column(Integer, primary_key=True)
client_id = Column(Integer, nullable=False)
supervisor_id = Column(Integer, nullable=False)
client = association_proxy("supervisor_association", "client")
engine = create_engine('sqlite:///temp.db', echo=True)
db_session = scoped_session(sessionmaker(bind=engine))
Base.metadata.create_all(bind=engine)
c1, c2 = Client(), Client()
sp1, sp2 = Supervisor(), Supervisor()
sch1, sch2, sch3 = Schedule(), Schedule(), Schedule()
sp1.clients = [c1]
c2.supervisors = [sp2]
c2.supervisor_associations[0].schedules = [sch1, sch2]
c1.supervisor_associations[0].schedules = [sch3]
db_session.add_all([c1, c2, sp1, sp2, ])
db_session.commit()
print c1.schedules
print sp2.schedules
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