在sqlalchemy中創建多對多關系時出錯

Question

我正在使用sqlalchemy文檔中的以下代碼（ http://docs.sqlalchemy.org/en/latest/orm/basic_relationships.html#many-to-many ）創建多對多表關系：

association_table = Table('association', Base.metadata,
    Column('left_id', Integer, ForeignKey('left.id')),
    Column('right_id', Integer, ForeignKey('right.id'))
)

class Parent(Base):
    __tablename__ = 'left'
    id = Column(Integer, primary_key=True)
    children = relationship(
        "Child",
        secondary=association_table,
        back_populates="parents")

class Child(Base):
    __tablename__ = 'right'
    id = Column(Integer, primary_key=True)
    parents = relationship(
        "Parent",
        secondary=association_table,
        back_populates="children")

運行應用程序時，出現以下錯誤：

sqlalchemy.exc.ProgrammingError: (psycopg2.ProgrammingError) relation "left" does not exist
 [SQL: '\nCREATE TABLE association_table (\n\tleft_id INTEGER, \n\tright_id INTEGER, \n\tFOREIGN KEY(left_id) REFERENCES left (id), \n\tFOREIGN KEY(right_id) REFERENCES right (id)\n)\n\n']

似乎是雞和雞蛋的問題-我似乎無法創建association_table，因為不存在Parent（即“左”）-但我不能先創建它，因為它稱為association_table。

有任何想法嗎？

Answer 1

嘗試在關系中使用表的字符串名稱。

class Parent(Base):
  __tablename__ = 'left'
  id = Column(Integer, primary_key=True)
  children = relationship(
    "Child",
    secondary='association',
    back_populates="parents")

class Child(Base):
  __tablename__ = 'right'
  id = Column(Integer, primary_key=True)
  parents = relationship(
    "Parent",
    secondary='association',
    back_populates="children")

association_table = Table('association', Base.metadata,
  Column('left_id', Integer, ForeignKey('left.id')),
  Column('right_id', Integer, ForeignKey('right.id'))
)

這樣，輔助聯接中的關聯表是從基本元數據表中獲取的，而不是順序執行查詢，我想這可能是錯誤的原因。