python - How to find the set of entities more recent than the last one with children -

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i have 2 sqlalchemy model objects designated thus:

class specinstance(base):     spec_id = column(integer, foreignkey('spec.spec_id'))     details = column(string)  class spec(base):     spec_id = column(integer)     spec_date = column(datetime)     instances = relationship(specinstance, backref="spec", cascade="all, delete, delete-orphan")

i looking query return spec objects have spec_date greater recent 1 instances. example, given objects these:

spec(spec_id=1, spec_date='2010-10-01') spec(spec_id=2, spec_date='2010-10-02') spec(spec_id=3, spec_date='2010-10-03')  specinstance(spec_id=2, details='whatever')

i want query return spec 3. spec 2 ineligible because has instances. spec 1 ineligible because it's older spec 2.

how do this?

i'm not testing code since i'm pretty sure work , setting env overhead.

in plain sql, 1 subquery. in sqlalchemy, subqueries created in manner:

sq = session.query(spec.spec_date.label('most_recent'))\             .join((specinstance, specinstance.spec_id==spec.spec_id))\             .order_by(desc(spec.spec_date))\             .limit(1).subquery()

here, joined 2 tables spec specinstances taken account, order them date latest on top, , take first - youngest instances - , need date. not executed - prepared subquery in:

session.query(spec)\        .join((sq, spec.spec_date>sq.c.most_recent))

which pretty straightforward. careful put double parentheses on join constructs, , include .c in second query on sq, since 'most_recent' dynamic column lookup.


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