(新的SQLAlchemy用户警报)我有三个表:一个人,该人从特定日期开始的每小时费率以及每日时间报告。我正在寻找一种正确的方法,以使该天的时基成本不高于该天的小时费率。
是的,我可以在创建时计算该值,并将其作为模型的一部分,但可以将其视为总结幕后更复杂数据的示例。如何计算Time.cost?它是hybrid_propery,column_property还是完全不同的东西?
class Person(Base):
__tablename__ = 'person'
personID = Column(Integer, primary_key=True)
name = Column(String(30), unique=True)
class Payrate(Base):
__tablename__ = 'payrate'
payrateID = Column(Integer, primary_key=True)
personID = Column(Integer, ForeignKey('person.personID'))
hourly = Column(Integer)
starting = Column(Date)
__tableargs__ =(UniqueConstraint('personID', 'starting',
name='uc_peron_starting'))
class Time(Base):
__tablename__ = 'entry'
entryID = Column(Integer, primary_key=True)
personID = Column(Integer, ForeignKey('person.personID'))
workedon = Column(Date)
hours = Column(Integer)
person = relationship("Person")
def __repr__(self):
return "<{date} {hours}hrs ${0.cost:.02f}>".format(self,
date=self.workedon.isoformat(), hours=to_hours(self.hours))
@property
def cost(self):
'''Cost of entry
'''
## This is where I am stuck in propery query creation
return self.hours * query(Payrate).filter(
and_(Payrate.personID==personID,
Payrate.starting<=workedon
).order_by(
Payrate.starting.desc())
最佳答案
为了尽可能优雅地解决问题,您这里遇到的问题使用了非常先进的SQLAlchemy技术,因此我知道您是一个初学者,但是此答案将向您展示到底。但是,解决这样的问题需要一次完成一个步骤,您可以通过不同的方式获得所需的答案。
在研究如何将其混合或混合之前,您需要考虑一下SQL。我们如何在任意一系列行中查询Time.cost?因为我们有一个简单的外键,所以我们可以干净地将“时间”链接到“人”。但是用这种特定的模式将时间链接到Payrate是很棘手的,因为Time不仅通过person_id而且还通过workon链接到Payrate-在SQL中,我们将使用“time.person_id = person.id AND time”最容易地加入到Payrate。在payrate.start_date和payrate.end_date之间工作”。但是您这里没有“end_date”,这意味着我们也必须派生该值。这个推导是最棘手的部分,所以我想出的是这样的(我将您的列名小写):
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
可能还有其他方法可以做到这一点,但这就是我想出的方式-几乎可以肯定,其他方式也会发生一些类似的事情(即子查询,联接)。
因此,通过开始/结束支付率,我们可以确定查询的外观。我们想使用BETWEEN将时间条目与日期范围匹配,但是最新的薪金率条目的“结束”日期将为NULL,因此一种解决方法是对非常高的日期使用COALESCE(另一种方法是使用条件):
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
现在,当在SQL表达式级别运行时,@ hybrid在SQLAlchemy中可以为您做的就是“entry.hours * payrate_derived.hourly”部分,仅此而已。所有的JOIN等都需要从外部提供给Hybrid。
因此,我们需要将大型子查询放入其中:
class Time(...):
@hybrid_property
def cost(self):
# ....
@cost.expression
def cost(cls):
return cls.hours * <SOMETHING>.hourly
因此,让我们弄清楚
<SOMETHING>
是什么。将那个SELECT建立为一个对象: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()
在表达方面,
cost()
混合体将需要引用payrate_derived(我们将在一分钟内完成python方面):class Time(...):
@hybrid_property
def cost(self):
# ....
@cost.expression
def cost(cls):
return cls.hours * payrate_derived.c.hourly
然后,为了使用我们的
cost()
混合,它必须在具有该联接的查询的上下文中。请注意,这里我们使用Python的datetime.date.max
来获取最大日期(方便!):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()
因此联接很大而且很笨拙,我们将需要经常这样做,更不用说在进行Python内混合时需要在Python中加载相同的集合。我们可以使用
relationship()
映射到它,这意味着我们必须设置自定义连接条件,但是我们还需要使用鲜为人知的非主映射器技术来实际映射到该子查询。非主映射器为您提供了一种将类映射到任意表或SELECT构造的方法,仅用于选择行。通常,我们永远不需要使用它,因为Query已经可以查询任意列和子查询,但是要从relationship()
中获取它,需要映射。映射需要定义一个主键,并且关系还需要知道关系的哪一侧是“外部”。这是这里最高级的部分,在这种情况下,它的工作方式如下: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
)
)
)
)
这是我们必须看到的该联接的最后一个。现在,我们可以更早地进行查询了:
print session.query(Person.name, Time.workedon, Time.hours, Time.cost).\
select_from(Time).\
join(Time.person).\
join(Time.payrate).\
all()
最后,我们还可以将新的
payrate
关系连接到Python级别的混合对象中: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
我们在这里采取的解决方案花费了很多精力,但是至少最复杂的部分(即薪金率映射)完全在一个地方,而且我们无需再去研究它。
这是一个完整的工作示例:
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
输出(第一行是聚合版本,其余是每个对象):
[(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