6
我想在sqlalchemy中生成这个查询。数据库中存在'demande'表。有一个使用generate_series函数生成时间步长的子查询。如何在sqlalchemy中生成此查询?
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;
我已经试过这样:
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
但它与InvalidRequesError抱怨:找不到FROM子句从加盟。我想这是由子查询造成的。
,如果我尝试“反转”的查询,它的工作原理,但它确实“LEFT OUTER JOIN”:
q = session.query(
func.count(Demande.id),
stmt.c.timestep).
outerjoin((stmt, grouped==stmt.c.timestep)).
group_by(stmt.c.timestep)
由于没有右外中SQLAlchemy的JOIN,我只是想找到一种方法将子查询作为第一个表格,将“demande”表格作为第二个表格。这样,我就可以使用LEFT OUTER JOIN
我要试试这个,明天在工作中,感谢你的回答,看起来不错 – 2010-10-25 19:53:01
好的,你已经学会了一些东西,但我仍然需要努力去实现我所期待的。不管怎么说,还是要谢谢你 – 2010-10-26 08:05:16