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我有一个多客户平台，可以有多个标签的人，并让一个带有一个标签的foo对象。我不希望每个客户的用户与其他客户分享他们的联系人（个人）或标签或foo。每个客户的每个标签名称都是唯一的带复合键的sqlalchemy辅助表
我怎么能对标签（名称，CUSTOMER_NAME）一复合键和人员之间的N-M的关系和标签？
我尝试这样做：
person_label_table = Table('person_label', Base.metadata,
Column('person_id', Integer, ForeignKey('person.id'), primary_key=True),
Column('name', Unicode(32), ForeignKey('label.name'), primary_key=True),
Column('customer_name', String(32), ForeignKey('label.customer_name'), primary_key=True)
)
class Person(Base, SaveMixin):
__tablename__ = 'person'
id = Column(Integer, primary_key=True)
labels = relationship('Label', secondary=person_label_table)
class Label(Base, SaveMixin):
__tablename__ = 'label'
name = Column(Unicode(32), primary_key=True)
customer_name = Column(String(32), ForeignKey('customer.name'), primary_key=True)
color = Column(String(32))
但我有此错误：
sqlalchemy.exc.AmbiguousForeignKeysError: Can't determine join between 'label' and 'person_label'; tables have more than one foreign key constraint relationship between them. Please specify the 'onclause' of this join explicitly.
我也试图与一个更经典的链接表person_label（label_id，为person_id）和添加一个标签的标签，但标签ID的必须加载到前台网络，当我做session.merge()与标签（没有ID）我有：
sqlalchemy.exc.IntegrityError: (sqlite3.IntegrityError) UNIQUE constraint failed: label.name, label.customer_name [SQL: 'INSERT INTO label (name, customer_name, color) VALUES (?, ?, ?)'] [parameters: ('foo', 'bar', 'grey')]
所以你会如何处理这种情况？
谢谢你的时间。
编辑： 万一它可以帮助时，布伦丹·阿贝尔响应，并显示代码后，我有这个错误：
sqlalchemy.exc.AmbiguousForeignKeysError: Could not determine join condition between parent/child tables on relationship Person.labels - there are multiple foreign key paths linking the tables via secondary table 'person_label'. Specify the 'foreign_keys' argument, providing a list of those columns which should be counted as containing a foreign key reference from the secondary table to each of the parent and child tables.
，所以我改变的定义标签亲自：
labels = relationship('Label', secondary=person_label_table, foreign_keys=[id]):
但我当时：
sqlalchemy.exc.NoForeignKeysError: Could not determine join condition between parent/child tables on relationship Person.labels - there are no foreign keys linking these tables via secondary table 'person_label'. Ensure that referencing columns are associated with a ForeignKey or ForeignKeyConstraint, or specify 'primaryjoin' and 'secondaryjoin' expressions.
我结束了（上添加人primaryjoin/secondaryjoin和消除对复合键的ForeignKey的）：
class Person(Base, SaveMixin):
__tablename__ = 'person'
labels = relationship('Label', secondary=person_label_table,
primaryjoin='and_(Person.id == person_label.c.person_id)',
secondaryjoin='and_(person_label.c.name == Label.name, person_label.c.customer_name == Label.customer_name)')
person_label_table = Table('person_label', Base.metadata,
Column('person_id', Integer, ForeignKey('person.id'), primary_key=True),
Column('name', Unicode(32), primary_key=True),
Column('customer_name', String(32), primary_key=True),
ForeignKeyConstraint(['name', 'customer_name'], ['label.name', 'label.customer_name'])
)