python - Sqlalchemy:FROM 中的子查询必须有一个别名

Question

我如何构造此 sqlalchemy 查询以使其执行正确的操作？

我已经给出了所有我能想到的别名，但我仍然得到:

ProgrammingError: (psycopg2.ProgrammingError) subquery in FROM must have an alias
LINE 4: FROM (SELECT foo.id AS foo_id, foo.version AS ...

此外，正如 IMSoP 指出的那样，它似乎试图将其变成交叉连接，但我只是希望它通过同一张表上的子查询将一个表与一个组连接起来。

这是 sqlalchemy:

(注意:我已将其重写为一个尽可能完整的独立文件，并且可以从 python shell 运行)

from sqlalchemy import create_engine, func, select
from sqlalchemy import Column, BigInteger, DateTime, Integer, String, SmallInteger
from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy.orm import sessionmaker

engine = create_engine('postgresql://postgres:#######@localhost:5435/foo1234')
session = sessionmaker()
session.configure(bind=engine)
session = session()

Base = declarative_base()

class Foo(Base):
     __tablename__ = 'foo'
     __table_args__ = {'schema': 'public'}
     id = Column('id', BigInteger, primary_key=True)
     time = Column('time', DateTime(timezone=True))
     version = Column('version', String)
     revision = Column('revision', SmallInteger)

foo_max_time_q = select([
     func.max(Foo.time).label('foo_max_time'),
     Foo.id.label('foo_id')
 ]).group_by(Foo.id
 ).alias('foo_max_time_q')

foo_q = select([
    Foo.id.label('foo_id'),
    Foo.version.label('foo_version'),
    Foo.revision.label('foo_revision'),
    foo_max_time_q.c.foo_max_time.label('foo_max_time')
]).join(foo_max_time_q, foo_max_time_q.c.foo_id == Foo.id
).alias('foo_q')

thing = session.query(foo_q).all()
print thing

生成的 sql:

SELECT foo_id AS foo_id,
    foo_version AS foo_version,
    foo_revision AS foo_revision,
    foo_max_time AS foo_max_time,
    foo_max_time_q.foo_max_time AS foo_max_time_q_foo_max_time,
    foo_max_time_q.foo_id AS foo_max_time_q_foo_id
    FROM (SELECT id AS foo_id,
        version AS foo_version,
        revision AS foo_revision,
        foo_max_time_q.foo_max_time AS foo_max_time
        FROM (SELECT max(time) AS foo_max_time,
            id AS foo_id GROUP BY id
        ) AS foo_max_time_q)
    JOIN (SELECT max(time) AS foo_max_time,
            id AS foo_id GROUP BY id
    ) AS foo_max_time_q
    ON foo_max_time_q.foo_id = id

这是玩具 table :

CREATE TABLE foo (
id bigint ,
time timestamp with time zone,
version character varying(32),
revision smallint
);

我期望得到的 SQL(期望的 SQL)是这样的:

SELECT foo.id AS foo_id,
       foo.version AS foo_version,
       foo.revision AS foo_revision,
       foo_max_time_q.foo_max_time AS foo_max_time
       FROM foo
       JOIN (SELECT max(time) AS foo_max_time,
            id AS foo_id GROUP BY id
            ) AS foo_max_time_q
        ON foo_max_time_q.foo_id = foo.id

最后的注释:如果可能的话，我希望使用 select() 而不是 session.query() 得到答案。谢谢

Answer 1

你快到了。做一个"selectable"子查询并通过 join() 将其与主查询连接起来:

foo_max_time_q = select([func.max(Foo.time).label('foo_max_time'),
                         Foo.id.label('foo_id')
                        ]).group_by(Foo.id
                         ).alias("foo_max_time_q")

foo_q = session.query(
          Foo.id.label('foo_id'),
          Foo.version.label('foo_version'),
          Foo.revision.label('foo_revision'),
          foo_max_time_q.c.foo_max_time.label('foo_max_time')
                ).join(foo_max_time_q, 
                       foo_max_time_q.c.foo_id == Foo.id)

print(foo_q.__str__())

打印(手动美化):

SELECT 
    foo.id AS foo_id, 
    foo.version AS foo_version, 
    foo.revision AS foo_revision, 
    foo_max_time_q.foo_max_time AS foo_max_time 
FROM 
    foo 
JOIN 
    (SELECT 
         max(foo.time) AS foo_max_time, 
         foo.id AS foo_id 
     FROM 
         foo 
     GROUP BY foo.id) AS foo_max_time_q 
ON 
    foo_max_time_q.foo_id = foo.id

此 gist 中提供了完整的工作代码.