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我有一個多客戶平臺，可以有多個標籤的人，並讓一個帶有一個標籤的foo對象。我不希望每個客戶的用戶與其他客戶分享他們的聯繫人（個人）或標籤或foo。每個客戶的每個標籤名稱都是唯一的帶複合鍵的sqlalchemy輔助表
我怎麼能對標籤（名稱，CUSTOMER_NAME）一複合鍵和人員之間的N-M的關係和標籤？
我嘗試這樣做：
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))
但我有此錯誤：
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.
我也試圖與一個更經典的鏈接表person_label（label_id，爲person_id）和添加一個標籤的標籤，但標籤ID的必須加載到前臺網絡，當我做session.merge()與標籤（沒有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')]
所以你會如何處理這種情況？
謝謝你的時間。
編輯： 萬一它可以幫助時，布倫丹·阿貝爾響應，並顯示代碼後，我有這個錯誤：
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.
，所以我改變的定義標籤親自：
labels = relationship('Label', secondary=person_label_table, foreign_keys=[id]):
但我當時：
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.
我結束了（上添加人primaryjoin/secondaryjoin和消除對複合鍵的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'])
)