Unable to create self referencing foreign key in flask-sqlalchemy

问题

I have a model Region and each Region can have sub-regions. Each sub-region has a field parent_id which is the id of its parent region. Here is how my model looks like

class Region(db.Model):
    __tablename__ = 'regions'
    __table_args__ = {'schema': 'schema_name'}
    id = db.Column(db.Integer, primary_key=True)
    name = db.Column(db.String(100))
    parent_id = db.Column(db.Integer, db.ForeignKey('regions.id'))
    parent = db.relationship('Region', primaryjoin=('Region.parent_id==Region.id'), backref='sub-regions')
    created_at = db.Column(db.DateTime, default=db.func.now())
    deleted_at = db.Column(db.DateTime)

Bu when i try to do db.create_all i get this error sqlalchemy.exc.NoReferencedTableError: Foreign key associated with column 'regions.parent_id' could not find table 'regions' with which to generate a foreign key to target column 'id'

Why cant it find regions when i am specifying it in __tablename__? I am using flask-sqlalchemy version 1.0

EDIT -- i removed the line

__table_args__ = {'schema': 'schema_name'}

from my code and it works. Beats the hell out of me.

回答1:

You must tell SQLAlchemy what the "remote side" of the relationship is, to distinguish between the current row and the row being joined to. The relevant explanation is located partway into this section of the documentation on relationships.

The relationship might look like this:

parent = db.relationship('Region', remote_side=id, backref='sub_regions')

This is an example demonstrating a self-referential relationship:

from sqlalchemy import Column, Integer, String, ForeignKey
from sqlalchemy.engine import create_engine
from sqlalchemy.ext.declarative.api import declarative_base
from sqlalchemy.orm import sessionmaker, relationship

engine = create_engine('sqlite:///:memory:', echo=True)
Session = sessionmaker(engine)
Base = declarative_base(engine)

session = Session()


class Region(Base):
    __tablename__ = 'region'

    id = Column(Integer, primary_key=True)
    name = Column(String, nullable=False)
    parent_id = Column(Integer, ForeignKey('region.id'), index=True)

    parent = relationship(lambda: Region, remote_side=id, backref='sub_regions')


Base.metadata.create_all()

r1 = Region(name='United States of America')
r2 = Region(name='California', parent=r1)

session.add_all((r1, r2))
session.commit()

ca = session.query(Region).filter_by(name='California').first()
print ca.parent.name

There will be many lines of SQL output since echo is on, and the script will print 'United States of America' at the end.

回答2:

I had the same issue with the schema name argument. What I changed to get it to work was to reference the table class directly in ForeignKey and relationships instead of using a string. Example:

parent_id = Column(Integer, ForeignKey(Region.id), index=True)

parent = relationship(lambda: Region, remote_side=id, backref='sub_regions')

回答3:

If you use a schema for any table, other tables that have foreign keys referencing those schema tables must provide the name of the schema. See the docs here

class Table(db.Model):
    __tablename__ = 'table_1_name'
    __table_args__ = {'schema': 'my_schema'}

    id = Column('id', Integer, primary_key=True)
    ...

class AnotherTable(db.Model):
    __tablename__ = 'table_2_name'
    # Doesn't matter if this belongs to the same or different schema
    # __table_args__ = {'schema': 'my_schema'}

    id = Column('id', Integer, primary_key=True)
    t1_id = Column(Integer, ForeignKey('my_schema.table_1_name.id'))
    ...

Works for both SQLAlchemy and Flask-SQLAlchemy. Hope this helps. :D

回答4:

I only see slight differences from @davidism, but here's what works for me in straight SQLAlchemy.

from sqlalchemy import Column
from sqlalchemy import Integer
from sqlalchemy import ForeignKey
from sqlalchemy.orm import relationship
from sqlalchemy.orm import backref

class Region(Base):
  __tablename__ = 'region'

  id = Column(Integer, primary_key=True)
  parent_id = Column(Integer, ForeignKey('region.id'), index=True)
  sub_regions = relationship('Region', backref=backref('parent', remote_side='Region.id'))

As he points out I'm guessing you wont need the imports, but should prefix them with db, so something like:

class Region(db.Model):
  __tablename__ = 'region'

  id = db.Column(db.Integer, primary_key=True)
  parent_id = db.Column(db.Integer, db.ForeignKey('region.id'), index=True)
  sub_regions = db.relationship('Region', backref=db.backref('parent', remote_side='Region.id'))

来源：https://stackoverflow.com/questions/20830147/unable-to-create-self-referencing-foreign-key-in-flask-sqlalchemy

标签

python

sqlalchemy

Flask

flask-sqlalchemy