SQLAlchemy 导入关系表

Question

I have problem with separating tables with relationships in different files. I want the tables below to be in three separate files and to import TableA in third party page, but I can not manage the load order.

In most of the time I'm receiving the following error.

sqlalchemy.exc. InvalidRequestError: When initializing mapper Mapper|TableA|tablea, expression 'TableB' failed to locate a name ("name 'TableB' is not defined"). If this is a class name, consider adding this relationship() to the class after both dependent classes have been defined.

class TableA(Base):
    __tablename__ = "tablea"
   id = Column(Integer, primary_key=True)
   name = Column(String)

   tableB = relationship("TableB", secondary = TableC.__table__)

class TableB(Base):
   __tablename__ = "tableb"
   id = Column(Integer, primary_key=True)
  name = Column(String)

class TableC(Base):
   __tablename__ = "tableab"
   tableAId = Column("table_a_id", Integer, ForeignKey("TableA.id"), primary_key=True)
   tableBId = Column("table_b_id", Integer, ForeignKey("TableB.id"), primary_key=True)

Answer 1

This should work (note that the TableC. table is replaced with the name of the table to avoid circular module loading):

### base.py
engine = create_engine('sqlite:///:memory:', echo=True)
Session = sessionmaker(bind=engine)
Base = declarative_base(bind=engine)

### classA.py
from base import Base
from classB import TableB

class TableA(Base):
    __tablename__ = 'tablea'
    id = Column(Integer, primary_key=True)
    name = Column(String(50))
    tableBs = relationship("TableB", secondary="tableab")
    #tableBs = relationship("TableB", secondary=TableC.__table__)

### classB.py
from base import Base

class TableB(Base):
    __tablename__ = 'tableb'
    id = Column(Integer, primary_key=True)
    name = Column(String(50))

### classC.py
from base import Base
from classA import TableA
from classB import TableB

class TableC(Base):
    __tablename__ = 'tableab'
    tableAId = Column(Integer, ForeignKey("tablea.id"), primary_key=True, )
    tableBId = Column(Integer, ForeignKey("tableb.id"), primary_key=True, )

### main.py
from base import Base, Session, engine
from classA import TableA
from classB import TableB
from classC import TableC
Base.metadata.create_all(engine)

Also I believe that the ForeignKey parameter is case sensitive, so you code might not work because "TableA.id" doe snot match "tablea" name when case-sensitive.

Answer 2

from sqlalchemy import Column, String, Integer
from sqlalchemy.ext.declarative import declarative_base
Base = declarative_base()


class Parent(Base):
    __tablename__ = 'Parent'
    ParentID = Column(Integer, primary_key=True)
    Description = Column(String)

    def __init__(self, ParentID, Description):
        self.ParentID = ParentID
        self.Description = Description
----------------------------------------------------------------------

from sqlalchemy import Column, String, Integer, ForeignKey
import Parent
from sqlalchemy.ext.declarative import declarative_base
Base = declarative_base()


class Child(Base):
    __tablename__ = "Child"
    ChildID = Column(Integer, primary_key=True)
    Description = Column(String)
    ParentID = Column('CompanyID', Integer, ForeignKey(Parent.ParentID))

    def __init__(self, ChildID, Description,ParentID):
        self.ChildID = ChildID
        self.Description = Description
        self.ParentID=ParentID

Answer 3

Simply modify the first line for 'ForeignKey' class and add another for the relationship class like below:

from sqlalchemy import Integer, ForeignKey, String, Column

from sqlalchemy.orm import relationship

To turn off the modification warnings, add another line

app.config['SQLALCHEMY_TRACK_MODIFICATIONS'] = True

Change the 'tableab' table name to 'tableb', this is the reason it is giving error that tableB is not found. Moreover, Python is case sensitive and we have to remember this.

Hope this will suffice.