Question

我使用SQLAlchemy的声明性API为我的应用程序构建我的模型和表。

但是，由于错误Base.metadata.create_all(engine)

，我无法使用

NoReferencedTableError: Foreign key associated with column 'org_prediction.company_id' could not find table 'organization' with which to generate a foreign key to target column 'id'

命令在我的数据库中构建新表

发生这种情况的原因是因为我试图为ForeignKey引用的类组织在models.py文件中不存在。所以，当我调用Base.metadata.create_all(engine)时，SQLAlchemy会因为类组织在Base中不存在而窒息......

我已经尝试反映我的数据库，它将返回存在的Organization的表，但是在models.py文件中没有为它编写相应的Class。但是当我反映数据库表并尝试在组织之上创建我的新依赖类时，我得到一个错误：InvalidRequestError: Table 'user' is already defined for this MetaData instance. Specify 'extend_existing=True' to redefine options and columns on an existing Table object.，这是因为反映的MetaData和我的models.py类用于用户重叠，以及SQLAlchemy不确定该怎么做。

但是我在哪里应用extend_existing = True？我没有Table对象来传递此参数。当我的models.py文件中不存在我的FK依赖表时，如何创建我的FK依赖表？或者我如何反映我现有的表，但只追加或更新数据库中尚不存在的类？

models.py

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

class User(Base):
    __tablename__ = 'user'
    id = Column(Integer, primary_key=True)
    username = Column(String(50))

class OrgPrediction(Base):
    __tablename__ = 'org_prediction'
    id = Column(Integer, primary_key=True)
    company_id = Column(String(255), ForeignKey('organization.id'), nullable=True)
    prediction = Column(String(255)
Base.metadata.create_all(engine)

NoReferencedTableError：与列'org_prediction.company_id'关联的外键无法找到用于生成目标列'id'的外键的表'organization'

MYSQL：

mysql> desc organization;
+-----------------------+------------+------+-----+---------+-------+
| Field                 | Type       | Null | Key | Default | Extra |
+-----------------------+------------+------+-----+---------+-------+
| id                 | bigint(20) | YES  | MUL | NULL    |       |


mysql> show tables;
+-------------------------+
| Tables_in_database |
+-------------------------+
| organization            |
| user                    |
| user_email              |

对于Reflection的models.py文件：

from sqlalchemy.ext.automap import automap_base
from sqlalchemy import create_engine

Base = automap_base()
Base.prepare(engine, reflect=True)

class User(Base):
    __tablename__ = 'user'
    id = Column(Integer, primary_key=True)
    username = Column(String(50))

class OrgPrediction(Base):
    __tablename__ = 'org_prediction'
    id = Column(Integer, primary_key=True)
    company_id = Column(String(255), ForeignKey('organization.id'), nullable=True)

Base.metadata.create_all(engine)

InvalidRequestError：已为此MetaData实例定义表'user'。指定'extend_existing = True'以重新定义现有Table对象上的选项和列。

Answer 1

有一种方法可以做到这一点，它涉及在添加Foreign-Key-dependency表之前使用Base.metadata.reflect（only = [＆＃39; table_name＆＃39;]）功能。我测试的最终代码：

Base = declarative_base()
class User(Base):
    __tablename__ = 'user'
    id = Column(Integer, primary_key=True)
    username = Column(String(50))

#see what tables exist in metadata right now
In [85]: Base.metadata.sorted_tables
Out[85]: [Table('user', MetaData(bind=None), Column('id', Integer(), table=<user>, primary_key=True, nullable=False), Column('username', String(length=50), table=<user>), schema=None)]

# this will add the organization table/class to the Base's Metadata. This also needs to happen before the Foreign Key reference in OrgPrediction class.
Base.metadata.reflect(bind=engine, only=['organization'])

# now this class' foreign key will reference organization.id which exists in Base metadata.
class OrgPrediction(Base):
    __tablename__ = 'org_prediction'
    id = Column(Integer, primary_key=True)
    company_id = Column(String(255), ForeignKey('organization.id'), nullable=True)
    prediction = Column(String(255))

# this should pass because the foreign key constraint is met.
Base.metadata.create_all(engine)

如何在SQLAlchemy 1.2.7中创建没有定义类的新表

1 个答案: