如何在sqlalchemy中生成此查询？

Question

我想在sqlalchemy中生成此查询.表'demande'存在于数据库中.有一个子查询使用generate_series函数生​​成时间步长.

SELECT
    timesteps.timestep AS timestep, d.count AS count
FROM
    (SELECT
        DATE_TRUNC('hour',date_demande) AS timestep,
        COUNT(id) AS count
     FROM
        demande
     GROUP BY
        timestep
    ) AS d

RIGHT OUTER JOIN
    (SELECT
        timestep
     FROM
        generate_series('2010-01-01 00:00:00'::timestamp,
                        '2010-01-01 23:59:59'::timestamp,
                        '1 hour'::interval) AS timestep
     ) AS timesteps
  ON d.timestep = timesteps.timestep

ORDER BY timestep;

我试过这个:

stmt = session.query(
        func.
            generate_series(
                datetime.datetime(2010,1,1,0,0,0),
                datetime.datetime(2010,1,1,23,59,59),
                cast('1 hour',Interval())).
            label('timestep')
        ).subquery()
print stmt
q = session.query(
        stmt.c.timestep,
        func.count(Demande.id)).
    outerjoin((Demande, grouped==stmt.c.timestep)).
    group_by(stmt.c.timestep)
print q

但它抱怨InvalidRequesError:无法找到要加入的FROM子句.我想这是由子查询引起的.

如果我试图"反转"查询,它可以工作,但它做'左外连接':

q = session.query(
        func.count(Demande.id),
        stmt.c.timestep).
    outerjoin((stmt, grouped==stmt.c.timestep)).
    group_by(stmt.c.timestep)

由于sqlalchemy中没有RIGHT OUTER JOIN,我只想找到一种方法将子查询作为第一个表,将'demande'表作为第二个表.这样我就可以使用LEFT OUTER JOIN了

Answer 1

下面的例子应该给你一个线索（假设我正确地猜测 Demande 是声明性模型）：

joined = stmt.outerjoin(Demande.__table__, Demande.grouped==stmt.c.timestep)
q = session.query(stmt.c.timestep, func.count(Demande.id)).\
        select_from(joined).\
        group_by(stmt.c.timestep)