Sqlalchemy get records with less than N parents - sqlalchemy

Question

How to get records with less than 3 parents or without parents?
Hello everyone! I have 3 tables: Product, Order and User. Product and Order, Product and User are linked many to many.
How to get products with less than 3 orders and users?
My query is:
stmt = select(User).where(Product.user_id == User.id).subquery()
stmt1 = select(Order).where(Order.user_id == User.id).subquery()
select(Product)
.outerjoin(Product.orders)
.outerjoin(Product.users)
filter(
and_(
select(func.count("*")).select_from(stmt).as_scalar() < 3,
select(func.count("*")).select_from(stmt1).as_scalar() <3,
)
)
If I remove the second condition from _and, I get 36 products, although I only have 20.
If I apply scalars().unique().all(), I get 20 products, although I should get less, because some of them already have 3 users or 3 orders.
With second _and condition I get 0 products
Thanks you in advance!

Answer 1

Since you have many-to-many relationships, I'll assume you must have at least association tables (if not association objects).
I have managed to extend from your query, using the association tables and it returns the expected results.
stmt = (
select(Product)
.outerjoin(Product.orders)
.outerjoin(Product.users)
.group_by(Product.id)
.having(
and_(
select(func.count())
.select_from(product_order_AT)
.where(product_order_AT.c.product_id == Product.id)
.as_scalar()
< 3,
select(func.count())
.select_from(product_user_AT)
.where(product_user_AT.c.product_id == Product.id)
.as_scalar()
< 3,
)
)
)
Now, it's not very pretty but it is readable, and produces the output you seem to want.
Full demo.
from sqlalchemy import (
Column,
ForeignKey,
Integer,
Table,
and_,
create_engine,
func,
select,
)
from sqlalchemy.orm import Session, declarative_base, relationship
Base = declarative_base()
product_order_AT = Table(
"product_order_AT",
Base.metadata,
Column("order_id", ForeignKey("order.id"), primary_key=True),
Column("product_id", ForeignKey("product.id"), primary_key=True),
)
product_user_AT = Table(
"product_user_AT",
Base.metadata,
Column("user_id", ForeignKey("user.id"), primary_key=True),
Column("product_id", ForeignKey("product.id"), primary_key=True),
)
class Product(Base):
__tablename__ = "product"
id = Column(Integer, primary_key=True)
orders = relationship(
"Order", secondary=product_order_AT, back_populates="products"
)
users = relationship(
"User", secondary=product_user_AT, back_populates="products"
)
def __repr__(self):
return f"Product(orders={self.orders}, users={self.users})"
class Order(Base):
__tablename__ = "order"
id = Column(Integer, primary_key=True)
products = relationship(
"Product", secondary=product_order_AT, back_populates="orders"
)
def __repr__(self):
return f"Order(id={self.id})"
class User(Base):
__tablename__ = "user"
id = Column(Integer, primary_key=True)
products = relationship(
"Product", secondary=product_user_AT, back_populates="users"
)
def __repr__(self):
return f"User(id={self.id})"
engine = create_engine("sqlite://", echo=True, future=True)
Base.metadata.create_all(engine)
with Session(engine) as session:
p1 = Product()
p2 = Product()
p3 = Product()
p4 = Product()
o1 = Order()
o2 = Order()
o3 = Order()
o4 = Order()
u1 = User()
u2 = User()
u3 = User()
u4 = User()
p1.orders.extend([o1, o2, o3, o4])
p1.users.extend([u1, u2, u3, u4])
p2.orders.extend([o1, o2, o3, o4])
p2.users.extend([u1, u2])
p3.orders.extend([o1, o2])
p3.users.extend([u1, u2, u3, u4])
p4.orders.extend([o1, o2])
p4.users.extend([u1, u2])
session.add_all([p1, p2, p3, p4])
session.commit()
with Session(engine) as session:
stmt = (
select(Product)
.outerjoin(Product.orders)
.outerjoin(Product.users)
.group_by(Product.id)
.having(
and_(
select(func.count())
.select_from(product_order_AT)
.where(product_order_AT.c.product_id == Product.id)
.as_scalar()
< 3,
select(func.count())
.select_from(product_user_AT)
.where(product_user_AT.c.product_id == Product.id)
.as_scalar()
< 3,
)
)
)
print(session.scalars(stmt).all())
prints
[Product(orders=[Order(id=2), Order(id=1)], users=[User(id=1), User(id=2)])]