Question

我有两个表Items和ItemDescriptions

class Item(Base):
    __tablename__ = "item"
    id = Column(Integer, primary_key=True)
    description_id = Column(Integer, ForeignKey('item_description.id'))

class ItemDescription(Base):
   __tablename__ = "item_description"
   id = Column(Integer, primary_key=True)

鉴于ItemDescriptions列表，我想要一个Items列表，以便每个ItemDescription id都有一个项目。我不关心哪个项目。

[为清晰起见而编辑]

给出这个项目和描述清单：

Item, Description
1   , 1
2   , 1
3   , 1
4   , 2
5   , 2
6   , 3
7   , 3
8   , 3

我想要一个返回类似的查询：

Item, Description
2   , 1
4   , 2
7   , 3

我在处理子查询等方面遇到了麻烦。

感谢您的帮助

Answer 1

我是column_property的忠实粉丝。以下是使用column_property执行所需操作的方法：

from sqlalchemy import *
from sqlalchemy.orm import *
from sqlalchemy.ext.declarative import declarative_base

Base = declarative_base()

class Item(Base):
    __tablename__ = 'item'

    id = Column(Integer, primary_key=True)
    description_id = Column(Integer, ForeignKey('item_description.id'))

class ItemDescription(Base):
    __tablename__ = 'item_description'

    id = Column(Integer, primary_key=True)

    any_item_id = column_property(
        select(
            [Item.id],
            id == Item.description_id,
            limit=1,
        ).label('any_item_id'),
        deferred=True,
    )

e = create_engine('sqlite://', echo=True)
Base.metadata.create_all(e)

s = Session(e)

descriptions = [
    ItemDescription(id=1),
    ItemDescription(id=2),
    ItemDescription(id=3),
]

s.add_all(descriptions)

items = [
    Item(id=1, description_id=1),
    Item(id=2, description_id=1),
    Item(id=3, description_id=1),
    Item(id=4, description_id=2),
    Item(id=5, description_id=2),
    Item(id=6, description_id=3),
    Item(id=7, description_id=3),
    Item(id=8, description_id=3),
]

s.add_all(items)

query = s.query(ItemDescription).options(undefer('any_item_id'))
for description in query:
    print description.any_item_id, description.id

# alternative way without using column_property
query = s.query(
    select(
        [Item.id],
        ItemDescription.id == Item.description_id,
        limit=1,
    ).label('any_item_id'),
    ItemDescription,
)
for item_id, description in query:
    print item_id, description.id

Answer 2

from sqlalchemy.orm import relationship, backref

class Item(Base):
    __tablename__ = "item"
    id = Column(Integer, primary_key=True)
    description_id = Column(Integer, ForeignKey('item_description.id'))
    desc = relationship(User, backref=backref('desc', order_by=id))

class ItemDescription(Base):
   __tablename__ = "item_description"
   id = Column(Integer, primary_key=True)

Now your every ItemDescription class will have an backref called `desc` which is nothing but a list of Item.

Now you can do something like this

item_desc = session.query(ItemDescription).\
...                        options(joinedload('desc').all()

for item in item_desc:
    print item.desc

我认为这可能不会给你确切的答案。它会帮助我猜测

如何根据SqlAlchemy中的表B选择表A中的一个？

2 个答案: