Python Sqlalchemy - tablename as a variable

Question

I am using SQLAlchemy in Python and am declaring my classes inheriting from a declarative base as follows:

from sqlalchemy import Column, Integer, String
from sqlalchemy.ext.declarative import declarative_base

Base = declarative_base()

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

As a user I would like to define the __tablename__ as a parameter, and not a hard-coded value , something like this:

class SomeClass(Base):
    __tablename__ = f'{environment}_some_table'
    id = Column(Integer, primary_key=True)
    name =  Column(String(50))

It is my understanding that f'{environment}_some_table' will be be evaluated when I import this package, and therefore I won't be able to modify it at a later stage (i.e. in an interactive notebook). I have a broken piece of code that tries to solve this through nested classes and encapsulation, but I do not manage to reference the instance variable of an outer class.

class Outer:
    def __init__(self, env):
        self.environment = env

    class SomeClass(Base):
        __tablename__ = f'{self.environment}_some_table'
        id = Column(Integer, primary_key=True)
        name =  Column(String(50))

I have read in a couple of SO questions that it is better not to use nested classes since no special relationship is established between these classes. So what kind of design should I use to solve this problem?

Thanks in advance!

"It is my understanding that f'{environment}_some_table' will be be evaluated before runtime" depends on what you mean by "runtime". In the usual sense of the word it is indeed evaluated at runtime. — Ilja Everilä
– Ilja Everilä, Commented Dec 11, 2017 at 13:24
Okay, I actually checked out what the definition of runtime is and you are right. I will modify the question to reflect this. Thanks! — Pezze
– Pezze, Commented Dec 11, 2017 at 15:43

Yoav Glazner · Accepted Answer · 2017-12-09 21:17:31Z

5

you can make all your model definitions inside a function scope so the will be depended on outer arguments:

def create_models(environment):
    class SomeClass(Base):
        __tablename__ = f'{environment}_some_table'
        id = Column(Integer, primary_key=True)
        name =  Column(String(50))
    ...
    globals().update(locals()) # update outer scope if needed

... # some time later
create_models('cheese')
... # now create the db session/engine/etc ...

another choice to look at is the builtin reload method. check it out :)

answered Dec 9, 2017 at 21:17

Yoav Glazner

8,0511 gold badge22 silver badges36 bronze badges

Sign up to request clarification or add additional context in comments.

Collectives™ on Stack Overflow

Python Sqlalchemy - tablename as a variable

1 Answer 1

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related