假设我的db模型包含一个对象User:
Base = declarative_base()
class User(Base):
__tablename__ = 'users'
id = Column(String(32), primary_key=True, default=...)
name = Column(Unicode(100))
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我的数据库包含一个包含n行的users表.在某些时候,我决定拆分成和,而在我想我的数据迁移以及.namefirstnamelastnamealembic upgrade head
自动生成的Alembic迁移如下:
def upgrade():
op.add_column('users', sa.Column('lastname', sa.Unicode(length=50), nullable=True))
op.add_column('users', sa.Column('firstname', sa.Unicode(length=50), nullable=True))
# Assuming that the two new columns have been committed and exist at
# this point, I would like to iterate over all rows of the name column,
# split the string, write it into the new …Run Code Online (Sandbox Code Playgroud) 我的模型看起来像
class Category(UserMixin, db.Model):
__tablename__ = 'categories'
uuid = Column('uuid', GUID(), default=uuid.uuid4, primary_key=True,
unique=True)
name = Column('name', String, nullable=False)
parent = Column('parent', String, nullable=False)
created_on = Column('created_on', sa.types.DateTime(timezone=True),
default=datetime.utcnow())
__table_args__ = (UniqueConstraint('name', 'parent'),)
def __init__(self, name, parent):
self.name = name
self.parent = parent
def __repr__(self):
return '<Category:%s:%s:%s>' % (
self.uuid, self.name, self.category_type)
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其中GUID是自定义sqlalchemy类型我使用alembic --autogenerate选项创建表
op.create_table('categories',
sa.Column('uuid', UUID(), nullable=False),
sa.Column('name', sa.String(), nullable=False),
sa.Column('parent', sa.String(), nullable=False),
sa.Column('created_on', sa.DateTime(timezone=True),
nullable=True),
sa.PrimaryKeyConstraint('uuid'),
sa.UniqueConstraint('name', 'parent'),
sa.UniqueConstraint('uuid')
)
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和PostgreSQL表一样
Table "public.categories"
Column …Run Code Online (Sandbox Code Playgroud) 我正在使用 alembic 来管理我的数据库结构。
添加使用 id 作为整数和主键的表后,id 列将成为自动增量列。我如何查询升级脚本中的数据,以便我确定我得到了正确的 id(我知道在这种特定情况下它是 1)?
我知道怎么做
#creating the table
op.create_table(
'srv_feed_return_type',
sa.Column('id', sa.Integer, primary_key=True),
sa.Column('name', sa.String(50), nullable=False),
sa.Column('created', sa.DateTime, server_default=func.now(), nullable=False),
sa.Column('created_by', sa.String(50), nullable=False),
sa.Column('last_updated', sa.DateTime, nullable=False),
sa.Column('last_updated_by', sa.String(50), nullable=False)
)
#table for operations
srv_feed_return_type = table('srv_feed_return_type',
column('name'),
column('created'),
column('created_by'),
column('last_updated'),
column('last_updated_by'))
#bulk insert
op.bulk_insert(srv_feed_return_type,
[
{'name': 'dataset',
'created': datetime.now(), 'created_by': 'Asken',
'last_updated': datetime.now(), 'last_updated_by': 'Asken'}
])
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我知道我可以进行更新,但是如何使用类似下面的内容进行选择?
op.execute(
srv_feed_return_type.update().\
where(srv_feed_return_type.c.name==op.inline_literal('dataset')).\
values({'name':op.inline_literal('somethingelse')})
)
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