如何根據SqlAlchemy中的表B選擇表A中的一個？

Question

我有兩個表項和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) 

鑑於ItemDescriptions我想這樣的項目有每個ItemDescription ID的一個項目的名單列表。我不在乎哪個項目。

[編輯爲清晰起見]

鑑於項目和說明的這個名單：

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

我想要一個查詢，將返回類似：

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

我無法完成子查詢等。

謝謝你的幫助

Answer 1

我是column_property的忠實粉絲。這裏有一種方法可以用column_property來做你想做的事情：

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 

我想這可能不會給你確切的答案。它會幫助我猜