What Type do I use for SQLalchemy declarative_base?

Question

I would like to use a Type for my SQLAlchemy base classes, but can't seem to find the type definitions.

Eg:

from sqlalchemy.types import BigInteger
from sqlalchemy.schema import Column, Index
from sqlalchemy.ext.declarative import declarative_base

Base = declarative_base()

class User(Base):

    __tablename__ = "users"

    id_user = Column(
        BigInteger, primary_key=True, autoincrement=True, nullable=False, index=True
    )
    username = Column(String, unique=True, index=True)

Now when referencing this User class, I don't know what type to use. Furthermore, If I wanted to have a dictionary which contains table_name : model mapping like:

from typing import Any, Dict

table_model_map: Dict[str, Any] = {
    "users": User,
    "another_table": AnotherTableModel,
}

How would I go about defining the types for this Dict

Thank you very much for helping me understand this!

dylanmorroll · Answer 1 · 2021-06-03T08:37:52.220

Ilja is right in that if all your model classes subclass Base, this is a suitable option for type hinting your dict.

If, however, you wanted to type hint something that receives Base as an argument, before you have created it, i.e. something like the following:

class DatabaseJanitor:
    def __init__(self, base: WhatTypeHere?, engine):
        self._base = base
        self._engine = engine

    def create_metadata(self):
        self._base.metadata.create_all(self._engine)

That is a bit more tricky... I did some digging and this is what I found for sqlalchemy-1.4.17.

The following can be seen in sqlalchemy/orm/decl_api.py:

class DeclarativeMeta(type):
    def __init__(cls, classname, bases, dict_, **kw):
        ...

def declarative_base(
    ...
    metaclass=DeclarativeMeta,
):
    ...
    return registry(
        _bind=bind,
        metadata=metadata,
        class_registry=class_registry,
        constructor=constructor,
    ).generate_base(
        mapper=mapper,
        cls=cls,
        name=name,
        metaclass=metaclass,
    )

class registry(object):
    ...
    def generate_base(
        ...
        metaclass=DeclarativeMeta,
    ):
        metadata = self.metadata

        bases = not isinstance(cls, tuple) and (cls,) or cls

        class_dict = dict(registry=self, metadata=metadata)
        if isinstance(cls, type):
            class_dict["__doc__"] = cls.__doc__

        if self.constructor:
            class_dict["__init__"] = self.constructor

        class_dict["__abstract__"] = True
        if mapper:
            class_dict["__mapper_cls__"] = mapper

        return metaclass(name, bases, class_dict)

So declarative_base will return an instance of the metaclass keyword argument passed in. So Base = declarative_base(metaclass=MyNewMetaclass) will result in type(Base) == MyNewMetaclass.

Furthermore, as you can see in registry.generate_base, all the attributes for the new instance of the metaclass (such as metadata) are contained in class_dict. These are implicitly added as attributes to the instance during creation, so they are not available during type hinting.

(Side note: I did a bit of research and it appears that because DeclarativeMeta subclasses type, type.__new__ is implicitly called which then adds the values in the class_dict dictionary as attributes to the instance of DeclarativeMeta. I'll link some resources for further reading at the end of this answer.)

So even if you type hint your function like so:


class DatabaseJanitor:
    def __init__(self, base: DeclarativeMeta, engine):
        self._base = base
        self._engine = engine

    def create_metadata(self):
        self._base.metadata.create_all(self._engine)

In create_metadata, your type hinting will warn Unresolved attribute reference 'metadata' for class 'DeclarativeMeta'.

So there's not really a good way to type hint declarative_base properly. My approach would be to just add it to the docstring for now.

Hope this helps for any future readers! :)

Here's some further reading on type/ metaclasses:

We can just do `type(declarative_base())` and it will point to `sqlalchemy.orm.decl_api.DeclarativeMeta`. But unfortunately it does not have proper type hint as you said. — jackblk, Feb 15 '22 at 08:05
This is very extensive and well written. Even though I'm disappointed with the outcome, you saved an hour of trying to understand the library's code (which is quite confusing). — miksus, Mar 13 '22 at 08:58
Looking at [the code](https://github.com/sqlalchemy/sqlalchemy/blob/9a8d039716068ded890e726ba344620907d86170/lib/sqlalchemy/orm/decl_api.py#L1183-L1189) the `generate_base()` function uses [`Any`](https://docs.python.org/3/library/typing.html#typing.Any) as the `metaclass` type and return type, with a default value [`DeclarativeMeta`](https://github.com/sqlalchemy/sqlalchemy/blob/9a8d039716068ded890e726ba344620907d86170/lib/sqlalchemy/orm/decl_api.py#L161-L163). I wonder if a more constrained [`TypeVar`](https://docs.python.org/3/library/typing.html#typing.TypeVar) would be useful here? — Jens, Jul 29 '22 at 00:00

score 2 · Answer 2 · answered Oct 17 '19 at 04:06

2

Base is a type that all your models (that are also types) inherit from, so

Dict[str, Base]

answered Oct 17 '19 at 04:06

Ilja Everilä

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What Type do I use for SQLalchemy declarative_base?

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