关于python：Sqlalchemy：FROM中的子查询必须具有别名

Sqlalchemy: subquery in FROM must have an alias

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，但是我仍然得到：

1 2	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所指出的那样，它似乎正在尝试将其转换为交叉联接，但我只希望它通过同一查询的子查询对同一张表进行联接。

这是sqlalchemy：

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

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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：

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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

这是玩具桌：

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

我期望获得的SQL(期望的SQL)将是这样的：

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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：
我希望尽可能使用select()而不是session.query()获得答案。谢谢