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POHow to generate this query in sqlalchemy?
 

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  1. POHow to generate this query in sqlalchemy?
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    <p>I want to generate this query in sqlalchemy. The table 'demande' exists in the database. There is a subquery that generates the timesteps with the generate_series function.</p> <pre><code>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; </code></pre> <p>I've tried this :</p> <pre><code>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 </code></pre> <p>But it complains with an InvalidRequesError: Could not find a FROM clause to join from. I guess this is caused by the subquery.</p> <p>If i try to "invert" the query, it works but it does a 'LEFT OUTER JOIN' :</p> <pre><code>q = session.query( func.count(Demande.id), stmt.c.timestep). outerjoin((stmt, grouped==stmt.c.timestep)). group_by(stmt.c.timestep) </code></pre> <p>As there is no RIGHT OUTER JOIN in sqlalchemy, I just want to find a way to take the subquery as the first table and the 'demande' table as the second one. This way I will be able to use LEFT OUTER JOIN</p>
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    1. USGhislain Leveque
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