在sqlalchemy中，如何仅使用表名快速定义相同的表模型（calss）是不同的？

Question

除了表名，我有10个表完全相同。而且我不想将10个表合并为一个。现在我尝试使用sqlalchemy将数据插入这些表中。我发现我在代码中重复执行以下操作：

class tableA(DeclarativeBase):
    __tablename__ = "A"
    id = Column(Integer, primary_key=True)
    title = Column('title', String(200))
    link = Column('link', String(200))

class tableB(DeclarativeBase):
    __tablename__ = "B"
    id = Column(Integer, primary_key=True)
    title = Column('title', String(200))
    link = Column('link', String(200))
    ......
    ......
class tableH(DeclarativeBase):
    __tablename__ = "H"
    id = Column(Integer, primary_key=True)
    title = Column('title', String(200))
    link = Column('link', String(200))

我可以使用表名作为输入的单个公共表类吗？

Answer 1

是的，有几种方法可以做到这一点，每种方法都有不同程度的冗长和灵活性：

Common Base Class

class Foo(DeclarativeBase):
    __abstract__ = True
    id = Column(Integer, primary_key=True)
    title = Column('title', String(200))
    link = Column('link', String(200))

class tableA(Foo):
    __tablename__ = "A"

密新

class Foo(object):
    id = Column(Integer, primary_key=True)
    title = Column('title', String(200))
    link = Column('link', String(200))

class tableA(Foo, DeclarativeBase):
    __tablename__ = "A"

动态type

def create_table(name):
    return type("table" + name, (DeclarativeBase,), {
        "__tablename__": name,
        "id": Column(Integer, primary_key=True),
        "title": Column('title', String(200)),
        "link": Column('link', String(200)),
    })

tableA = create_table("A")