2012-01-21 116 views
4

我想建立另一个许多一对多关系的关系,代码如下所示:SQLAlchemy的关系,许多一对多关联表

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) 

我想要做的是有关一个特定的客户 - 主管关系的时间表,但我还没有发现如何去做。通过SQLAlchemy文档,我发现了一些提示,导致Schedule-Table上的ForeignKeyConstraint。

如何指定关系以使此关联有效?

+0

什么是具有独立supervision_association_table,然后有一个一对一的关系给它的时间表类的目的是什么?为什么不使用一个表,然后使用关联对象模式? http://www.sqlalchemy.org/docs/orm/relationships.html#association-object。在任何情况下,如果您保留两个独立的“时间表”和“监督”表,您需要在这里映射supervision_associaiton_table以创建一系列关系链,然后再与相关项目建立关系链。 – zzzeek

+1

客户真的可以有超过1个主管?您的M-N表格建议*是*,但关系(主管)的名称意味着*否*。 – van

+0

@zzzeek我希望课程表类与监督有一对多的关系,这样一对客户主管可以有许多不同的时间表。 – dasmaze

回答

7

您需要映射supervision_association_table,以便您可以创建与其之间的关系。

我可能会在这里夸耀一些东西,但它似乎是因为你有多对多在这里你真的不能有Client.schedules - 如果我说Client.schedules.append(some_schedule),哪一行在“监督”是指向?所以下面的例子为那些加入每个SupervisorAssociation的Schedule集合的人提供了一个只读的“汇总”访问器。当方便时,association_proxy扩展用于隐藏SupervisionAssociation对象的细节。

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 
+0

这基本上就是我所说的,在关联表上有时间表并将它与一个类进行映射。我没有使用关联代理,尽管我可能会这样做。这次真是万分感谢。 – dasmaze