Python 如何编写带有连接和聚合的SQLAlchemy查询?

Python 如何编写带有连接和聚合的SQLAlchemy查询?,python,sqlalchemy,Python,Sqlalchemy,我有一个表,它有3列:type、content和time一个整数。对于每个“类型”,我要选择具有最大最近“时间”整数和相应数据的条目。如何使用SQLAlchemy和Python实现这一点?我可以使用SQL执行以下操作: select c.type, c.time, b.data from parts as b inner join (select a.type, max(a.time) as time from parts as a group b

我有一个表,它有3列:type、content和time一个整数。对于每个“类型”,我要选择具有最大最近“时间”整数和相应数据的条目。如何使用SQLAlchemy和Python实现这一点?我可以使用SQL执行以下操作:

select
  c.type,
  c.time,
  b.data
from
  parts as b

inner join

  (select
    a.type,
    max(a.time) as time
  from parts as a
  group by a.type) as c

on

b.type = c.type and
b.time = c.time
但是我怎样才能在炼金术中做到这一点呢

表映射:

class Structure(Base):
    __tablename__ = 'structure'
    id = Column(Integer, primary_key=True)
    type = Column(Text)
    content = Column(Text)
    time = Column(Integer)

    def __init__(self, type, content):
        self.type = type
        self.content = content
        self.time = time.time()

    def serialise(self):
        return {"type" : self.type,
            "content" : self.content};
尝试的查询:

    max = func.max(Structure.time).alias("time")
    c = DBSession.query(max)\
        .add_columns(Structure.type, Structure.time)\
        .group_by(Structure.type)\
        .subquery()
    c.alias("c")

    b = DBSession.query(Structure.content)\
        .add_columns(c.c.type, c.c.time)\
        .join(c, Structure.type == c.c.type)
给我:

sqlalchemy.exc.OperationalError:OperationalError靠近:语法 错误u'选择structure.content作为structure\u content,anon\u 1.type作为 anon_1_类型,anon_1.t时间作为anon_1_时间\n从结构联接选择 time.max_1为max_1,structure.type为type,structure.time为time \nFROM maxstructure.time AS time,structu re GROUP BY structure.type为structure.type=anon_1.type'


我基本上是在黑暗中刺伤,所以任何帮助都将不胜感激

使用子查询尝试下面的代码:

subq = (session.query(
            Structure.type, 
            func.max(Structure.time).label("max_time")
        ).
        group_by(Structure.type)
        ).subquery()

qry = (session.query(Structure).
       join(subq, and_(Structure.type == subq.c.type, Structure.time == subq.c.max_time))
       )

print qry
生成SQL:

SELECT  structure.id AS structure_id, structure.type AS structure_type, structure.content AS structure_content, structure.time AS structure_time
FROM    structure 
JOIN    (SELECT structure.type AS type, max(structure.time) AS max_time
         FROM structure GROUP BY structure.type) AS anon_1 
    ON  structure.type = anon_1.type 
    AND structure.time = anon_1.max_time

这可以通过使用SQLAlchemy来实现。你已经试过什么了吗?你有定义的表映射吗?@MartijnPieters嘿,Martijn,我昨晚在离开办公室的路上发布了这个。一旦我有机会再看一眼,我会给它添加一些代码;谢谢。@MartijnPieters到目前为止我已经添加了我的工作,但不知道我是否在正确的轨道上。赢家!感谢您的帮助,您已经澄清了我的主要问题以及一些较小的问题,如“我如何加入多个专栏?”,非常感谢您的帮助;谢谢