Question

我有一个Alchemy ORM对象：

from sqlalchemy.ext.declarative import declarative_base
Base = declarative_base()


class MyORM(Base):
    id = Column(Integer, primary_key=True)
    name = Column(String(128), unique=True, nullable=False)

使用alembic创建表格时，我会执行以下操作：

def upgrade():
    op.create_table(
        'myorm',
        sa.Column('id', sa.Integer, primary_key=True),
        sa.Column('name', sa.String(128), nullable=False),
    )

问题：有没有办法使用MyORM类创建表？像这样：

def upgrade():
    op.create_table(
        'myorm',
        sa.BaseObject(MyORM)
    )

Answer 1

这正是Alembic迁移试图避免的。如果将迁移绑定到模型的当前状态，则它将不是一致的升级路径。

您可以在迁移中使用声明来创建表并迁移数据，但不能更改。您将不得不重新创建与应用程序定义分开的定义。如果您想要进行数据迁移并且更熟悉ORM查询而不是核心查询，这将非常有用。

以下是一个示例迁移，它使用声明式创建具有多对多关系的Foo和Bar模型，创建表并插入一些数据。

"""declarative

Revision ID: 169ad57156f0
Revises: 29b4c2bfce6d
Create Date: 2014-06-25 09:00:06.784170
"""

revision = '169ad57156f0'
down_revision = '29b4c2bfce6d'

from alembic import op
import sqlalchemy as sa
from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy.orm import sessionmaker, relationship

Session = sessionmaker()
Base = declarative_base()


class Foo(Base):
    __tablename__ = 'foo'

    id = sa.Column(sa.Integer, primary_key=True)
    name = sa.Column(sa.String, nullable=False, unique=True)


class Bar(Base):
    __tablename__ = 'bar'

    id = sa.Column(sa.Integer, primary_key=True)
    name = sa.Column(sa.String, nullable=False, unique=True)
    foos = relationship(Foo, lambda: foo_bar, backref='bars')


foo_bar = sa.Table(
    'foo_bar', Base.metadata,
    sa.Column('foo_id', sa.Integer, sa.ForeignKey('foo.id'), primary_key=True),
    sa.Column('bar_id', sa.Integer, sa.ForeignKey('bar.id'), primary_key=True)
)

def upgrade():
    bind = op.get_bind()

    Base.metadata.create_all(bind=bind)

    session = Session(bind=bind)
    session._model_changes = False  # if you are using Flask-SQLAlchemy, this works around a bug

    f1 = Foo(name='f1')
    f2 = Foo(name='f2')
    b1 = Bar(name='b1')
    b2 = Bar(name='b2')

    f1.bars = [b1, b2]
    b2.foos.append(f2)

    session.add_all([f1, f2, b1, b2])
    session.commit()


def downgrade():
    bind = op.get_bind()

    # in this case all we need to do is drop the tables
    # Base.metadata.drop_all(bind=bind)

    # but we could also delete data
    session = Session(bind=bind)
    session._model_changes = False  # if you are using Flask-SQLAlchemy, this works around a bug

    b1 = session.query(Bar).filter_by(name='b1').one()

    session.delete(b1)
    session.commit()

使用declarative_base派生对象的alembic create_table

1 个答案: