SQLAlchemy cannot find a class name
PythonSqlalchemyPyramidRelationshipsPython Problem Overview
Simplified, I have the following class structure (in a single file):
Base = declarative_base()
class Item(Base):
__tablename__ = 'item'
id = Column(BigInteger, primary_key=True)
# ... skip other attrs ...
class Auction(Base):
__tablename__ = 'auction'
id = Column(BigInteger, primary_key=True)
# ... skipped ...
item_id = Column('item', BigInteger, ForeignKey('item.id'))
item = relationship('Item', backref='auctions')
I get the following error from this:
sqlalchemy.exc.InvalidRequestError
InvalidRequestError: When initializing mapper Mapper|Auction|auction, expression
'Item' failed to locate a name ("name 'Item' is not defined"). If this is a
class name, consider adding this relationship() to the Auction class after
both dependent classes have been defined.
I'm not sure how Python cannot find the Item class, as even when passing the class, rather than the name as a string, I get the same error. I've been struggling to find examples of how to do simple relationships with SQLAlchemy so if there's something fairly obvious wrong here I apologise.
Python Solutions
Solution 1 - Python
This all turned out to be because of the way I've set SQLAlchemy up in Pyramid. Essentially you need to follow this section to the letter and make sure you use the same declarative_base
instance as the base class for each model.
I was also not binding a database engine to my DBSession
which doesn't bother you until you try to access table metadata, which happens when you use relationships.
Solution 2 - Python
if it's a subpackage class, add Item
and Auction
class to __init__.py
in the subpackage.
Solution 3 - Python
The SQLAlchemy documentation on Importing all SQLAlchemy Models states in part:
> However, due to the behavior of SQLAlchemy's "declarative" configuration mode, all modules which hold active SQLAlchemy models need to be imported before those models can successfully be used. So, if you use model classes with a declarative base, you need to figure out a way to get all your model modules imported to be able to use them in your application.
Once I imported all of the models (and relationships), the error about not finding the class name was resolved.
- Note: My application does not use Pyramid, but the same principles apply.
Solution 4 - Python
I've solved the same error by inheriting a 'db.Model' instead of 'Base'... but I'm doing the flask
Eg:
from flask_sqlalchemy import SQLAlchemy
db = SQLAlchemy()
class someClass(db.Model):
someRelation = db.relationship("otherClass")
Solution 5 - Python
Also, even though this doesn't apply to the OP, for anyone landing here having gotten the same error, check to make sure that none of your table names have dashes in them.
For example, a table named "movie-genres" which is then used as a secondary in a SQLAlchemy relationship will generate the same error "name 'movie' is not defined"
, because it will only read as far as the dash. Switching to underscores (instead of dashes) solves the problem.
Solution 6 - Python
Case with me
Two models defined in separate files, one is Parent
and the other is Child
, related with a Foreign Key. When trying to use Child
object in celery, it gave
sqlalchemy.exc.InvalidRequestError: When initializing mapper Mapper|Child|child, expression 'Parent' failed to locate a name ("name 'Parent' is not defined"). If this is a class name, consider adding this relationship() to the <class 'app.models.child'>
parent.py
from app.models import *
class Parent(Base):
__tablename__ = 'parent'
id = Column(BigInteger, primary_key=True, autoincrement=True)
name = Column(String(60), nullable=False, unique=True)
number = Column(String(45), nullable=False)
child.py
from app.models import *
class Child(Base):
__tablename__ = 'child'
id = Column(BigInteger, primary_key=True, autoincrement=True)
parent_id = Column(ForeignKey('parent.id'), nullable=False)
name = Column(String(60), nullable=False)
parent = relationship('Parent')
Solution
Add an import statement for Parent
in beginning of child.py
child.py (modified)
from app.models import *
from app.models.parent import Parent # import Parent in child.py 👈👈
class Child(Base):
__tablename__ = 'child'
id = Column(BigInteger, primary_key=True, autoincrement=True)
parent_id = Column(ForeignKey('parent.id'), nullable=False)
name = Column(String(60), nullable=False)
parent = relationship('Parent')
Why this worked
The order in which models get loaded is not fixed in SQLAlchemy.
So, in my case, Child
was being loaded before Parent
. Hence, SQLAlchemy can't find what is Parent
. So, we just imported Parent
before Child
gets loaded.
Namaste
Solution 7 - Python
My Solution
One models file, or even further, if you need.
models.py
from sqlalchemy import Boolean, BigInteger, Column, DateTime, Float, ForeignKey, BigInteger, Integer, String
from sqlalchemy.orm import relationship
from sqlalchemy.ext.declarative import declarative_base
Base = declarative_base()
from .parent import Parent
from .child import Child
parent.py
from sqlalchemy import Boolean, BigInteger, Column, DateTime, Float, ForeignKey, BigInteger, Integer, String
from sqlalchemy.orm import relationship
from sqlalchemy.ext.declarative import declarative_base
#Base = declarative_base()
class Parent(Base):
__tablename__ = 'parent'
id = Column(BigInteger, primary_key=True, autoincrement=True)
name = Column(String(60), nullable=False, unique=True)
number = Column(String(45), nullable=False)
child.py
from sqlalchemy import Boolean, BigInteger, Column, DateTime, Float, ForeignKey, BigInteger, Integer, String
from sqlalchemy.orm import relationship
from sqlalchemy.ext.declarative import declarative_base
Base = declarative_base()
class Child(Base):
__tablename__ = 'child'
id = Column(BigInteger, primary_key=True, autoincrement=True)
parent_id = Column(ForeignKey('parent.id'), nullable=False)
name = Column(String(60), nullable=False)
parent = relationship('Parent')
Why this worked
Same Deepam answer, but with just one models.py file to import another models