SQLAlchemy 1.4 table query that includes first result of second table [duplicate] - sqlalchemy

Question

I have a SQLAlchemy model with a one-to-many relationship between table x and table y. The record (if any) with the greatest id in table y where y.x_id = x.id is special. Class X and class Y map tables x and y.
I know how to define X.all_y (ORDER BY y.id). How do I define X.latest_y equivalent to X.all_y[-1]?

Answer 1

the purely relational way to do it requires using a subquery to get the "latest" or "max" value, correlated to the parent, then equating that with the members of the collection. It means you'll get best results if you put an index on the column that determines "the latest":
from sqlalchemy import *
from sqlalchemy.orm import *
engine = create_engine('sqlite:///:memory:', echo='debug')
m = MetaData()
parent = Table('parent', m,
Column('id', Integer, primary_key=True)
)
child = Table('child', m,
Column('id', Integer, primary_key=True),
Column('parent_id', Integer, ForeignKey('parent.id')),
Column('sortkey', Integer)
)
m.create_all(engine)
class Parent(object):
def __init__(self, children):
self.all_c = children
class Child(object):
def __init__(self, sortkey):
self.sortkey = sortkey
latest_c = select([func.max(child.c.sortkey)]).\
where(child.c.parent_id==parent.c.id).\
correlate(parent).\
as_scalar()
mapper(Parent, parent, properties={
'all_c':relation(Child),
'latest_c':relation(Child,
primaryjoin=and_(
child.c.sortkey==latest_c,
child.c.parent_id==parent.c.id
),
uselist=False
)
})
mapper(Child, child)
session = sessionmaker(engine)()
p1, p2, p3 = Parent([Child('a'), Child('b'), Child('c')]), \
Parent([Child('b'), Child('c')]),\
Parent([Child('f'), Child('g'), Child('c')])
session.add_all([p1, p2, p3])
session.commit()
assert p1.latest_c.sortkey == 'c'
assert p2.latest_c.sortkey == 'c'
assert p3.latest_c.sortkey == 'g'
Alternatively, you can on some platforms use LIMIT, which can produce faster results since you avoid the aggregation and can join the collection item on its primary key:
latest_c = select([child.c.id]).\
where(child.c.parent_id==parent.c.id).\
order_by(child.c.sortkey.desc()).\
limit(1).\
correlate(parent).\
as_scalar()
mapper(Parent, parent, properties={
'all_c':relation(Child),
'latest_c':relation(Child,
primaryjoin=and_(
child.c.id==latest_c,
child.c.parent_id==parent.c.id
),
uselist=False
)
})

Answer 2

Here is a version of #zzzeek's answer that uses declarative rather than imperative mapping, using declared_attr to insert the relationship into the parent's __mapper_args__.
import sqlalchemy as sa
from sqlalchemy import orm
Base = orm.declarative_base()
class Child(Base):
__tablename__ = 'children'
id = sa.Column(sa.Integer, primary_key=True)
sortkey = sa.Column(sa.Integer, nullable=False)
parent_id = sa.Column(sa.Integer, sa.ForeignKey('parents.id'))
parent = orm.relationship('Parent', back_populates='children')
class Parent(Base):
__tablename__ = 'parents'
id = sa.Column(sa.Integer, primary_key=True)
children = orm.relationship('Child', back_populates='parent')
#orm.declared_attr
def __mapper_args__(cls):
children = Child.__table__
most_recent_child = (
sa.select(children.c.id)
.where(children.c.parent_id == cls.id)
.order_by(children.c.sortkey.desc())
.limit(1)
.correlate(cls.__table__)
.scalar_subquery()
)
rel = orm.relation(
Child,
primaryjoin=sa.and_(
Child.id == most_recent_child, Child.parent_id == cls.id
),
uselist=False,
viewonly=True,
)
return {'properties': {'latest_child': rel}}
# Build and test.
engine = sa.create_engine('sqlite://', echo=True, future=True)
Base.metadata.create_all(engine)
Session = orm.sessionmaker(engine, future=True)
with Session.begin() as s:
children = [Child(sortkey=i) for i in range(1, 6)]
parent = Parent(children=children)
s.add(parent)
with Session() as s:
w = s.scalars(sa.select(Parent)).first()
assert w.latest_child.sortkey == 5, f'{w.latest_child.sortkey=}'
assert len(w.children) == 5, f'{len(w.children)=}'