SQLAlchemy导入具有关系的表

Question

我在使用不同文件中的关系分隔表时遇到问题.我希望下面的表格在三个单独的文件中,并在第三方页面中导入TableA,但我无法管理加载顺序.

在大多数情况下,我收到以下错误.

sqlalchemy.exc. InvalidRequestError:初始化映射器Mapper | TableA | tablea时,表达式"TableB"无法找到名称("未定义名称'TableB'").如果这是类名,请考虑在定义了两个依赖类之后将此关系()添加到类中.

class TableA(Base):
    __tablename__ = "tablea"
   id = Column(Integer, primary_key=True)
   name = Column(String)

   tableB = relationship("TableB", secondary = TableC.__table__)

class TableB(Base):
   __tablename__ = "tableb"
   id = Column(Integer, primary_key=True)
  name = Column(String)

class TableC(Base):
   __tablename__ = "tableab"
   tableAId = Column("table_a_id", Integer, ForeignKey("TableA.id"), primary_key=True)
   tableBId = Column("table_b_id", Integer, ForeignKey("TableB.id"), primary_key=True)

Answer 1

这应该工作(注意TableC.表被替换为表的名称,以避免循环模块加载):

### base.py
engine = create_engine('sqlite:///:memory:', echo=True)
Session = sessionmaker(bind=engine)
Base = declarative_base(bind=engine)

### classA.py
from base import Base
from classB import TableB

class TableA(Base):
    __tablename__ = 'tablea'
    id = Column(Integer, primary_key=True)
    name = Column(String(50))
    tableBs = relationship("TableB", secondary="tableab")
    #tableBs = relationship("TableB", secondary=TableC.__table__)

### classB.py
from base import Base

class TableB(Base):
    __tablename__ = 'tableb'
    id = Column(Integer, primary_key=True)
    name = Column(String(50))

### classC.py
from base import Base
from classA import TableA
from classB import TableB

class TableC(Base):
    __tablename__ = 'tableab'
    tableAId = Column(Integer, ForeignKey("tablea.id"), primary_key=True, )
    tableBId = Column(Integer, ForeignKey("tableb.id"), primary_key=True, )

### main.py
from base import Base, Session, engine
from classA import TableA
from classB import TableB
from classC import TableC
Base.metadata.create_all(engine)

另外我认为ForeignKey参数区分大小写,因此您的代码可能不起作用,因为"TableA.id"在区分大小写时会与"tablea"名称匹配.