SqlAlchemy - Filtering by Relationship Attribute

Question

I don't have much experience with SQLAlchemy and I have a problem, which I can't solve. I tried searching and I tried a lot of code. This is my Class (reduced to the most significant code):

class Patient(Base):
    __tablename__ = 'patients'
    id = Column(Integer, primary_key=True, nullable=False)
    mother_id = Column(Integer, ForeignKey('patients.id'), index=True)
    mother = relationship('Patient', primaryjoin='Patient.id==Patient.mother_id', remote_side='Patient.id', uselist=False)
    phenoscore = Column(Float)

and I would like to query all patients, whose mother's phenoscore is (for example) == 10

As told, I tried a lot of code, but I don't get it. The logically solution, in my eyes, would be

patients = Patient.query.filter(Patient.mother.phenoscore == 10)

because, you can access .mother.phenoscore for each element when outputting but, this code doesn't do it.

Is there a (direct) possibility to filter by an attribute of a relationship (without writing the SQL Statement, or an extra join-statement), I need this kind of filter more than one time.

Even if there is no easy solution, I am happy to get all answers.

Solution 1

Use method has() of relationship (more readable):

patients = Patient.query.filter(Patient.mother.has(phenoscore=10))

or join (usually faster):

patients = Patient.query.join(Patient.mother, aliased=True)\
                    .filter_by(phenoscore=10)

Solution 2

You have to query the relationsip with join

You will get the example from this Self-Referential Query Strategies

Solution 3

Good news for you: I recently made package that gives you filtering/sorting with "magical" strings as in Django, so you can now write something like

Patient.where(mother___phenoscore=10)

It's a lot shorter, especially for complex filters, say,

Comment.where(post___public=True, post___user___name__like='Bi%')

Hope you will enjoy this package

https://github.com/absent1706/sqlalchemy-mixins#django-like-queries

Solution 4

I used it with sessions, but an alternate way where you can access the relationship field directly is

db_session.query(Patient).join(Patient.mother) \
    .filter(Patient.mother.property.mapper.class_.phenoscore==10)

I have not tested it, but I guess this would also work

Patient.query.join(Patient.mother) \
    .filter(Patient.mother.property.mapper.class_.phenoscore==10)

Solution 5

This is a more general answer on how to query relationships.

relationship(..., lazy='dynamic', ...)

This allows you to:

parent_obj.some_relationship.filter(ParentClass.some_attr==True).all()

View more solutions

Comments

[user1105851]

patients = Patient.query.filter(Patient.mother.has(Patient.phenoscore==‌​10))

[Denis Otkidach]

@user1105851 has() supports both condition expression as unnamed argument and filter_by-style keyword arguments. The later seems more readable to me.

[aruisdante]

@DenisOtkidach correct, but then it would be phenoscore = 10. filter_by only takes equality keywords (since it's just doing **kwargs on them)

[Denis Otkidach]

@aruisdante You are right, it was erroneous edit of the answer.

[Boston Kenne]

use any instead: patients = Patient.query.filter(Patient.mother.any(phenoscore=10))

[Ricky Levi]

it gives me: loaders cannot be used with many-to-one/one-to-one relationships and/or uselist=False