在sqlalchemy中创建多对多关系时出错

Question

我正在使用sqlalchemy文档中的以下代码（ http://docs.sqlalchemy.org/en/latest/orm/basic_relationships.html#many-to-many ）创建多对多表关系：

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")

运行应用程序时，出现以下错误：

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']

似乎是鸡和鸡蛋的问题-我似乎无法创建association_table，因为不存在Parent（即“左”）-但我不能先创建它，因为它称为association_table。

有任何想法吗？

Answer 1

尝试在关系中使用表的字符串名称。

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'))
)

这样，辅助联接中的关联表是从基本元数据表中获取的，而不是顺序执行查询，我想这可能是错误的原因。