Question

如何按日期连接sqlalchemy个无关系的查询和sort？

e.g。这些是模型：

Human
Car
Tree

他们都有列created。以下是查询：

q1 = session.query(Human.created.label('created'), Human).filter(...)
q2 = session.query(Car.created.label('created'), Car).filter(...)
q3 = session.query(Tree.created.label('created'), Tree).filter(...)

现在我想连接这3个查询和order_by日期。预期的结果将是，如：

date      | instance
----------------------------
<created> | Human<instance>
<created> | Car<instance>
<created> | Car<instance>
<created> | Tree<instance>
<created> | Human<instance>
...

Answer 1

根据提供的信息，我想要在不同的表上执行联合（https://www.w3schools.com/sql/sql_union.asp）。

在下面的示例中，创建了不同的表，并建立了这些不同表的并集。

from sqlalchemy import Column, Integer, String, DateTime
from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy import create_engine
from sqlalchemy.orm import sessionmaker
from datetime import datetime

Base = declarative_base()


class Human(Base):
    __tablename__ = 'human'
    id = Column('id', Integer, primary_key=True)
    created = Column('created', DateTime)
    name = Column('name', String(250))

    def __init__(self, created, name):
        self.created = created
        self.name = name

    def __repr__(self):
        return '<{created} - {name}>'.format(created=self.created, name=self.name)


class Car(Base):
    __tablename__ = 'car'
    id = Column('id', Integer, primary_key=True)
    created = Column('created', DateTime)
    brand = Column(String(250))

    def __init__(self, created, brand):
        self.created = created
        self.brand = brand

    def __repr__(self):
        return '<{created} - {brand}>'.format(created=self.created, brand=self.brand)


class Tree(Base):
    __tablename__ = 'tree'
    id = Column('id', Integer, primary_key=True)
    created = Column('created', DateTime)
    type = Column(String(250))

    def __init__(self, created, type):
        self.created = created
        self.type = type

    def __repr__(self):
        return '<{created} - {type}>'.format(created=self.created, type=self.type)


engine = create_engine('sqlite:///')
session = sessionmaker()
session.configure(bind=engine)
ex_ses = session()
Base.metadata.create_all(engine)

human = Human(datetime.now(), 'Human a')
human2 = Human(datetime.now(), 'Human b')
car = Car(datetime.now(), 'Car a')
car2 = Car(datetime.now(), 'Car b')
tree = Tree(datetime.now(), 'Tree a')
tree2 = Tree(datetime.now(), 'Tree b')

ex_ses.add(human)
ex_ses.add(human2)
ex_ses.add(car)
ex_ses.add(car2)
ex_ses.add(tree)
ex_ses.add(tree2)
ex_ses.commit()

# Query
h = ex_ses.query(Human.created, 'name')
c = ex_ses.query(Car.created, 'brand')
t = ex_ses.query(Tree.created, 'type')

print(h.union_all(c).union_all(t).all())

最终查询（在print语句中）返回：

[(datetime.datetime(2017, 7, 5, 5, 55, 54, 633863), 'Human a'),
(datetime.datetime(2017, 7, 5, 5, 55, 54, 633863), 'Human b'),
(datetime.datetime(2017, 7, 5, 5, 55, 54, 633863), 'Car a'),
(datetime.datetime(2017, 7, 5, 5, 55, 54, 633863), 'Car b'),
(datetime.datetime(2017, 7, 5, 5, 55, 54, 633863), 'Tree a'),
(datetime.datetime(2017, 7, 5, 5, 55, 54, 633863), 'Tree b')]

此答案基于：How to union across multiple tables in SQLAlchemy?

如何在SqlAlchemy

1 个答案: