Question

我正在python中构建一个应用程序。我正在使用sqlalchemy-migrate来跟踪我的数据库架构。我有一个表user_category，它有两列：id和name。我正在尝试使用user_category表的外键创建一个用户表。我创建用户表的更改脚本如下：

from sqlalchemy import *
from migrate import *
from migrate.changeset import *

meta = MetaData()
user_category = Table('user_category', meta)

user = Table('user', meta,
    Column('id', Integer, primary_key=True),
    Column('email', String(255)),
    Column('first_name', String(40)),
    Column('surname', String(40)),
    Column('password', String(255)),
    Column('user_category', Integer, ForeignKey("user_category.id")),
)

def upgrade(migrate_engine):
    # Upgrade operations go here. Don't create your own engine; bind migrate_engine
    # to your metadata
    meta.bind = migrate_engine
    user.create()

def downgrade(migrate_engine):
    # Operations to reverse the above upgrade go here.
    meta.bind = migrate_engine
    user.drop()

当我运行'manage.py test'时，我得到了错误：

sqlalchemy.exc.NoReferencedColumnError: Could not create ForeignKey 'user_catego
ry.id' on table 'user': table 'user_category' has no column named 'id'

Answer 1

可以告诉SQLAlchemy从数据库中自动加载表结构，而不是复制粘贴user_category表的定义：

from sqlalchemy import *
from migrate import *
from migrate.changeset import *

meta = MetaData()

user = Table('user', meta,
    Column('id', Integer, primary_key=True),
    Column('email', String(255)),
    Column('first_name', String(40)),
    Column('surname', String(40)),
    Column('password', String(255)),
    Column('user_category', Integer, ForeignKey("user_category.id")),
)

def upgrade(migrate_engine):
    _t = sa.Table('user_category', meta, autoload=True)
    meta.bind = migrate_engine
    user.create()

对不起，迟到的答案：）

Answer 2

您声称您的“user_category”表具有名称和ID。 “user_category”的定义根本不包含任何列：）

使用SQLAlchemy-Migrate创建具有外键约束的表时出错

2 个答案: