sql - 如何在 sqlalchemy 中生成此查询？

Question

我想在 sqlalchemy 中生成这个查询。数据库中存在表“demande”。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;

我已经试过了:

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

但它会提示 InvalidRequesError:无法找到要加入的 FROM 子句。我猜这是由子查询引起的。

如果我尝试“反转”查询，它会起作用，但它会执行“LEFT OUTER JOIN”:

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，所以我只想想办法把子查询作为第一个表，'demande'表作为第二个。这样我就可以使用 LEFT OUTER JOIN

Answer 1

下面的例子应该会给你一个线索(假设我猜对了 Demande 是声明式模型):

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)