SQLAlchemy relationship to any class - python

Is it possible to build a SQLAlchemy relationship that will accept any instance that inherits from the same Base? Ideally, it maintains one foreign key for the instance's ID and a second foreign key for the instance's type.
As an example, this Variable class has a relationship applies_to between itself and the Car class.
# some overhead, for completeness
from sqlalchemy import Column, String, Integer, ForeignKey
from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy.orm import relationship, Session
from sqlalchemy import create_engine
engine = create_engine('sqlite:///:memory:')
session = Session(bind=engine)
Base = declarative_base()
# defining my classes
class Variable(Base):
__tablename__ = 'variables'
id = Column(Integer, primary_key=True)
name = Column(String)
value = Column(Integer)
applies_to_id = Column(Integer, ForeignKey('cars.id'))
applies_to = relationship('Car')
class Car(Base):
__tablename__ = 'cars'
id = Column(Integer, primary_key=True)
class Truck(Base):
__tablename__ = 'trucks'
id = Column(Integer, primary_key=True)
Base.metadata.create_all(engine)
With it, I can make a Car and a Variable for that car and commit to the database.
car = Car()
var1 = Variable(name="speed", value=50, applies_to=car)
session.add(car)
session.add(var1)
session.commit()
But I cannot do a similar thing with Truck and Variable.
truck = Truck()
var2 = Variable(name="speed", value=50, applies_to=truck)
session.add(truck)
session.add(var2)
session.commit()
AssertionError: Attribute 'applies_to' on class '<class '__main__.Variable'>' doesn't handle objects of type '<class '__main__.Truck'>'
I would like to edit the Variable class to support Truck and any other classes that inherit from Base through the applies_to relationship.
EDIT: Similar to sqlalchemy generic foreign key (like in django ORM), but that solution requires all classes to inherit from the HasAddresses mixin and handles a backref via addresses on those classes. Is there a solution that only edits the Variable class and doesn't maintain a backref to the other classes?

Related

SQLAlchemy can't set schema on an inherited class using __table_args__

I am not able to define a specific schema on an inherited class using __table_args__ because SQLAlchemy throws sqlalchemy.exc.ArgumentError: Can't place __table_args__ on an inherited class with no table
I am aware of this question but this workaround does not work for defining the schema.
import sqlalchemy as sa
import sqlalchemy.ext.declarative
from sqlalchemy.ext.declarative import declarative_base
Base = declarative_base()
class User(Base):
__tablename__ = 'user'
id = sa.Column(sa.Integer, primary_key=True)
full_name = sa.Column(sa.Text, nullable=False)
class Employee(User):
company_id = sa.Column(sa.Integer)
__table_args__ = {'schema': 'foo'}
I'd like to declare Employee in schema foo
Do somebody know any workarounds?

Is it possible to rename the metadata attribute of a SQLAlchemy declarative base?

I'm trying to set up a database with a few specific fields (and I can't move away from the specification). One of the fields would be a column called metadata, but sqlalchemy prevents that:
sqlalchemy.exc.InvalidRequestError: Attribute name 'metadata' is reserved for the MetaData instance when using a declarative base class.
Is there a decent workaround for this? Do I need to monkeypatch the declarative_base function to rename the metadata attribute? I couldn't find an option to rename that attribute in the api docs.
Here's some example code that will fail with the above error:
#!/usr/bin/env python3.7
from sqlalchemy.ext.declarative import declarative_base, declared_attr
from sqlalchemy import Column, Integer
class CustomBase(object):
#declared_attr
def __tablename__(cls):
return cls.__name__.lower()
DBBase = declarative_base(cls=CustomBase)
class Data(DBBase):
id = Column(Integer, primary_key=True)
metadata = Column(Integer)
if __name__ == "__main__":
print(dir(Data()))

You can use like:
class Data(DBBase):
id = Column(Integer, primary_key=True)
# metadata = Column(Integer)
metadata_ = Column("metadata", Integer)
The constructor of Column class has a name parameter. You can find it from https://docs.sqlalchemy.org/en/13/core/metadata.html#sqlalchemy.schema.Column
The name field may be omitted at construction time and applied later
In other words, you could write a name as you want originally.

Python+SQLAlchemy: Getting mapper class from InstrumentedList?

I want to dynamically get the SQLAlchemy mapper class that instances of InstrumentedList use.
I have a 1-M ParentClass-ChildClass (let's call the column myRelation), and the parentInstance.myRelation are InstrumentedList instances. I could hack it and grab the class of the first instance in InstrumentedList, but this doesn't work if there are no objects in the InstrumentedList.
The reason: I need to append a Python dictionary containing the properties of mapper class X to the InstrumentedList, but I don't know the mapper class at runtime. Since I can't append a dict, I need to get the mapper class.
Thanks.

the relationship is associated with the instrumented list via property. so start with this plain vanilla mapping:
from sqlalchemy import *
from sqlalchemy.orm import *
from sqlalchemy.ext.declarative import declarative_base
Base = declarative_base()
class A(Base):
__tablename__ = "a"
id = Column(Integer, primary_key=True)
bs = relationship("B")
class B(Base):
__tablename__ = "b"
id = Column(Integer, primary_key=True)
a_id = Column(Integer, ForeignKey('a.id'))
in any recent version of SQLAlchemy you can see it like this:
print A.bs.property.mapper.class_
in 0.8 there's a little more API available and you can do this:
from sqlalchemy import inspect
print inspect(A.bs).mapper.class_
docs for inspect-> http://docs.sqlalchemy.org/en/rel_0_8/core/inspection.html
docs for "mapper->class_" -> http://docs.sqlalchemy.org/en/rel_0_8/orm/mapper_config.html#sqlalchemy.orm.mapper.Mapper.class_

One-to-many relationship to multiple models

I have a model Thing and a model Action. There is a one-to-many relationship between Things and Actions. However, I would like to be able to subclass Action to have (for example) BuildAction, HealAction and BloodyStupidAction. Is it possible using Flask-SQLAlchemy to do this and maintain the single one-to-many relationship?

This problem is described in the SQLAlchemy docs under Inheritance Configuration. If your different subclasses will share the same database table, you should use single table inheritance.
Code example:
class Thing(db.Model):
__tablename__ = 'thing'
id = db.Column(db.Integer, primary_key=True)
actions = db.relationship('Action', backref=db.backref('thing'))
class Action(db.Model):
__tablename__ = 'action'
id = db.Column(db.Integer, primary_key=True)
thing_id = db.Column(db.Integer, db.ForeignKey('thing.id'))
discriminator = db.Column('type', db.String(50))
__mapper_args__ = {'polymorphic_on': discriminator}
class BuildAction(Action):
__mapper_args__ = {'polymorphic_identity': 'build_action'}
time_required = db.Column(db.Integer)
Each subclass of Action should inherit the thing relationship defined in the parent class. The action.type column describes which subclass action each row of the table represents.

SQLAlchemy cannot find a class name

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.

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.

if it's a subpackage class, add Item and Auction class to __init__.py in the subpackage.

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.

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 🙏

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")

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.

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

I had a different error, but the answers in here helped me fix it.
The error I received:
sqlalchemy.exc.InvalidRequestError: When initializing mapper mapped class Parent->parents, expression 'Child' failed to locate a name ('Child'). If this is a class name, consider adding this relationship() to the <class 'parent.Parent'> class after both dependent classes have been defined.
My set-up is similar toDeepam's answer.
Briefly what I do different:
I have multiple separate .py files for each db.Model.
I use a construct/fill database .py file that pre-fills db.Model objects in either Multi-threading or single threading way
What caused the error:
Only in multi-threaded set up the error occured
This construct/fill .py script did import Parent, but not Child.
What fixed it:
Adding an import to Child fixed it.

I had yet another solution, but this helped clue me in. I was trying to implement versioning, from https://docs.sqlalchemy.org/en/14/orm/examples.html#versioning-objects using the "history_mapper" class.
I got this same error. All I had to do to fix it was change the order in which my models were imported.

Use back_populates for relationship mapping in both models.
Also keep in mind to import both the models in the models/__init__.py
Base = declarative_base()
class Item(Base):
__tablename__ = 'item'
id = Column(BigInteger, primary_key=True)
# ... skip other attrs ...
auctions = relationship('Auction', back_populates='item')
class Auction(Base):
__tablename__ = 'auction'
id = Column(BigInteger, primary_key=True)
# ... skipped ...
item_id = Column('item', BigInteger, ForeignKey('item.id'))
item = relationship('Item', back_populates='auctions')

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