The problem you have here, to solve as elegantly as possible, uses very advanced SQLAlchemy techniques, so I know you're a beginner, but this answer is going to show you all the way out to the end. However, solving a problem like this requires walking through one step at a time, and you can get the answer you want in different ways as we go through.
Before you get into how to hybrid this or whatever, you need to think about the SQL. How can we query for Time.cost over an arbitrary series of rows? We can link Time to Person cleanly because we have a simple foreign key. But to link Time to Payrate, with this particular schema is tricky, because Time links to Payrate not just via person_id but also via workedon - in SQL we'd join to this most easily using "time.person_id = person.id AND time.workedon BETWEEN payrate.start_date AND payrate.end_date". But you don't have an "end_date" here, which means we have to derive that also. That derivation is the trickiest part, so what I came up with starts like this (I've lowercased your column names):
SELECT payrate.person_id, payrate.hourly, payrate.starting, ending.ending
FROM payrate LEFT OUTER JOIN
(SELECT pa1.payrate_id, MIN(pa2.starting) as ending FROM payrate AS pa1
JOIN payrate AS pa2 ON pa1.person_id = pa2.person_id AND pa2.starting > pa1.starting
GROUP BY pa1.payrate_id
) AS ending ON payrate.payrate_id=ending.payrate_id
There might be other ways to get this, but that's what I came up with - other ways would almost certainly have some similar kind of thing going on (i.e. subqueries, joins).
So with a payrate starting/ending, we can figure out what a query would look like. We want to use BETWEEN to match a time entry to the date range, but the latest payrate entry will have NULL for the "end" date, so one way to work around that is to use COALESCE against a very high date (the other is to use conditionals):
SELECT *, entry.hours * payrate_derived.hourly
FROM entry
JOIN
(SELECT payrate.person_id, payrate.hourly, payrate.starting, ending.ending
FROM payrate LEFT OUTER JOIN
(SELECT pa1.payrate_id, MIN(pa2.starting) as ending FROM payrate AS pa1
JOIN payrate AS pa2 ON pa1.person_id = pa2.person_id AND pa2.starting > pa1.starting
GROUP BY pa1.payrate_id
) AS ending ON payrate.payrate_id=ending.payrate_id) as payrate_derived
ON entry.workedon BETWEEN payrate_derived.starting AND COALESCE(payrate_derived.ending, "9999-12-31")
AND entry.person_id=payrate_derived.person_id
ORDER BY entry.person_id, entry.workedon
Now what @hybrid can do for you in SQLAlchemy, when run at the SQL expression level, is exactly just the "entry.hours * payrate_derived.hourly" part, that's it. All the JOIN and such there, you'd need to provide externally to the hybrid.
So we need to stick that big subquery into this:
class Time(...):
@hybrid_property
def cost(self):
# ....
@cost.expression
def cost(cls):
return cls.hours * <SOMETHING>.hourly
So let's figure out what <SOMETHING> is. Build up that SELECT as an object:
from sqlalchemy.orm import aliased, join, outerjoin
from sqlalchemy import and_, func
pa1 = aliased(Payrate)
pa2 = aliased(Payrate)
ending = select([pa1.payrate_id, func.min(pa2.starting).label('ending')]).\
select_from(join(pa1, pa2, and_(pa1.person_id == pa2.person_id, pa2.starting > pa1.starting))).\
group_by(pa1.payrate_id).alias()
payrate_derived = select([Payrate.person_id, Payrate.hourly, Payrate.starting, ending.c.ending]).\
select_from(outerjoin(Payrate, ending, Payrate.payrate_id == ending.c.payrate_id)).alias()
The cost() hybrid, on the expression side, would need to refer to payrate_derived (we'll do the python side in a minute):
class Time(...):
@hybrid_property
def cost(self):
# ....
@cost.expression
def cost(cls):
return cls.hours * payrate_derived.c.hourly
Then in order to use our cost() hybrid, it would have to be in the context of a query that has that join. Note here we use Python's datetime.date.max to get that max date (handy!):
print session.query(Person.name, Time.workedon, Time.hours, Time.cost).\
select_from(Time).\
join(Time.person).\
join(payrate_derived,
and_(
payrate_derived.c.person_id == Time.person_id,
Time.workedon.between(
payrate_derived.c.starting,
func.coalesce(
payrate_derived.c.ending,
datetime.date.max
)
)
)
).\
all()
So that join is big, and klunky, and we will need to do it often, not to mention we're going to need to load that same collection in Python when we do our in-Python hybrid. We can map to it using relationship(), which means we have to set up custom join conditions, but also we need to actually map to that subquery, using a lesser-known technique called a non-primary mapper. A non-primary mapper gives you a way to map a class to some arbitrary table or SELECT construct just for the purposes of selecting rows. We normally never need to use this because Query already lets us query for arbitrary columns and subqueries, but to get it out of a relationship() it needs a mapping. The mapping needs a primary key to be defined, and the relationship also needs to know which side of the relationship is "foreign". This is the most advanced part here and in this case it works out like this:
from sqlalchemy.orm import mapper, relationship, foreign
payrate_derived_mapping = mapper(Payrate, payrate_derived, non_primary=True,
primary_key=[
payrate_derived.c.person_id,
payrate_derived.c.starting
])
Time.payrate = relationship(
payrate_derived_mapping,
viewonly=True,
uselist=False,
primaryjoin=and_(
payrate_derived.c.person_id == foreign(Time.person_id),
Time.workedon.between(
payrate_derived.c.starting,
func.coalesce(
payrate_derived.c.ending,
datetime.date.max
)
)
)
)
So that's the last we'd have to see of that join. We can now do our query earlier as:
print session.query(Person.name, Time.workedon, Time.hours, Time.cost).\
select_from(Time).\
join(Time.person).\
join(Time.payrate).\
all()
and finally we can wire up our new payrate relationship into the Python-level hybrid as well:
class Time(Base):
# ...
@hybrid_property
def cost(self):
return self.hours * self.payrate.hourly
@cost.expression
def cost(cls):
return cls.hours * payrate_derived.c.hourly
The solution we have here took a lot of effort, but at least the most complex part, that payrate mapping, is entirely in just one place and we never need to look at it again.
Here's a full working example:
from sqlalchemy import create_engine, Column, Integer, ForeignKey, Date, \
UniqueConstraint, select, func, and_, String
from sqlalchemy.orm import join, outerjoin, relationship, Session, \
aliased, mapper, foreign
from sqlalchemy.ext.declarative import declarative_base
import datetime
from sqlalchemy.ext.hybrid import hybrid_property
Base = declarative_base()
class Person(Base):
__tablename__ = 'person'
person_id = Column(Integer, primary_key=True)
name = Column(String(30), unique=True)
class Payrate(Base):
__tablename__ = 'payrate'
payrate_id = Column(Integer, primary_key=True)
person_id = Column(Integer, ForeignKey('person.person_id'))
hourly = Column(Integer)
starting = Column(Date)
person = relationship("Person")
__tableargs__ =(UniqueConstraint('person_id', 'starting',
name='uc_peron_starting'))
class Time(Base):
__tablename__ = 'entry'
entry_id = Column(Integer, primary_key=True)
person_id = Column(Integer, ForeignKey('person.person_id'))
workedon = Column(Date)
hours = Column(Integer)
person = relationship("Person")
@hybrid_property
def cost(self):
return self.hours * self.payrate.hourly
@cost.expression
def cost(cls):
return cls.hours * payrate_derived.c.hourly
pa1 = aliased(Payrate)
pa2 = aliased(Payrate)
ending = select([pa1.payrate_id, func.min(pa2.starting).label('ending')]).\
select_from(join(pa1, pa2, and_(
pa1.person_id == pa2.person_id,
pa2.starting > pa1.starting))).\
group_by(pa1.payrate_id).alias()
payrate_derived = select([Payrate.person_id, Payrate.hourly, Payrate.starting, ending.c.ending]).\
select_from(outerjoin(Payrate, ending, Payrate.payrate_id == ending.c.payrate_id)).alias()
payrate_derived_mapping = mapper(Payrate, payrate_derived, non_primary=True,
primary_key=[
payrate_derived.c.person_id,
payrate_derived.c.starting
])
Time.payrate = relationship(
payrate_derived_mapping,
viewonly=True,
uselist=False,
primaryjoin=and_(
payrate_derived.c.person_id == foreign(Time.person_id),
Time.workedon.between(
payrate_derived.c.starting,
func.coalesce(
payrate_derived.c.ending,
datetime.date.max
)
)
)
)
e = create_engine("postgresql://scott:tiger@localhost/test", echo=False)
Base.metadata.drop_all(e)
Base.metadata.create_all(e)
session = Session(e)
p1 = Person(name='p1')
session.add(p1)
session.add_all([
Payrate(hourly=10, starting=datetime.date(2013, 5, 17), person=p1),
Payrate(hourly=15, starting=datetime.date(2013, 5, 25), person=p1),
Payrate(hourly=20, starting=datetime.date(2013, 6, 10), person=p1),
])
session.add_all([
Time(person=p1, workedon=datetime.date(2013, 5, 19), hours=10),
Time(person=p1, workedon=datetime.date(2013, 5, 27), hours=5),
Time(person=p1, workedon=datetime.date(2013, 5, 30), hours=5),
Time(person=p1, workedon=datetime.date(2013, 6, 18), hours=12),
])
session.commit()
print session.query(Person.name, Time.workedon, Time.hours, Time.cost).\
select_from(Time).\
join(Time.person).\
join(Time.payrate).\
all()
for time in session.query(Time):
print time.person.name, time.workedon, time.hours, time.payrate.hourly, time.cost
Output (first line is the aggregate version, remainder is the per-object):
[(u'p1', datetime.date(2013, 5, 19), 10, 100), (u'p1', datetime.date(2013, 5, 27), 5, 75), (u'p1', datetime.date(2013, 5, 30), 5, 75), (u'p1', datetime.date(2013, 6, 18), 12, 240)]
p1 2013-05-19 10 10 100
p1 2013-05-27 5 15 75
p1 2013-05-30 5 15 75
p1 2013-06-18 12 20 240
