SQLAlchemy Expression Language problem - postgresql

Question

I'm trying to convert this to something sqlalchemy expression
language compatible, I don't know if it's possible out of box and
are hoping someone more experienced can help me along. The backend
is PostgreSQL and if I can't make it as an expression I'll create
a string instead::
SELECT
DISTINCT date_trunc('month', x.x) as date,
COALESCE(b.res1, 0) AS res1,
COALESCE(b.res2, 0) AS res2
FROM
generate_series(
date_trunc('year', now() - interval '1 years'),
date_trunc('year', now() + interval '1 years'),
interval '1 months'
) AS x
LEFT OUTER JOIN(
SELECT
date_trunc('month', access_datetime) AS when,
count(NULLIF(resource_id != 1, TRUE)) AS res1,
count(NULLIF(resource_id != 2, TRUE)) AS res2
FROM tracking_entries
GROUP BY
date_trunc('month', access_datetime)
) AS b
ON (date_trunc('month', x.x) = b.when)
First of all I got a class TrackingEntry mapped to tracking_entries,
the select statement within the outer joined can be converted to something
like (pseudocode)::
from sqlalchemy.sql import func, select
from datetime import datetime, timedelta
stmt = select([
func.date_trunc('month', TrackingEntry.resource_id).label('when'),
func.count(func.nullif(TrackingEntry.resource_id != 1, True)).label('res1'),
func.count(func.nullif(TrackingEntry.resource_id != 2, True)).label('res2')
],
group_by=[func.date_trunc('month', TrackingEntry.access_datetime), ])
Considering the outer select statement I have no idea how to build it, my guess is
something like::
outer = select([
func.distinct(func.date_trunc('month', ?)).label('date'),
func.coalesce(?.res1, 0).label('res1'),
func.coalesce(?.res2, 0).label('res2')
],
from_obj=[
func.generate_series(
datetime.now(),
datetime.now() + timedelta(days=365),
timedelta(days=1)
).label(x)
])
Then I suppose I have to link those statements together without using
foreign keys::
outer.outerjoin(stmt???).??(func.date_trunc('month', ?.?), ?.when)
Anyone got any suggestions or even better a solution?
http://pastie.org/994367

Answer 1

not a full answer but when you want to join select() objects together they have a ".c." attribute, they effectively gain the same interface as the Table itself. so
s1 = select(...)
s2 = select(...)
s3 = select([s1,s2]).select_from(s1.join(s2, s1.c.foo==s2.c.bar))
The same applies for functions
select ([func.foo(s1.c.x)]).select_from(s1.join(s2, ...))