nested sqlalchemy filter with parent and son - sql

Question

With the following scheme:
class User(Base):
id = Column(Integer, primary_key=True)
name = Column(String)
class Photo(Base):
id = Column(Integer, primary_key=True)
user_id = Column(Integer, ForeignKey(User.id), nullable=False)
user = relationship(User)
class Tag(Base):
id = Column(Integer, primary_key=True)
tag_name = Column(String)
tag_version = Column(Integer)
photo_id = Column(Integer, ForeignKey(Photo.id), nullable=False)
photo = relationship(Photo)
How do I create an SQLAlchemy query to get all the photos of a specific user, that don't have a specific tag and version.
As in "all the photos of the user with id "1234" that don't have a "cat" of version "2" tagged in them".
Also interesting would be "all the users who have at least one photo without a specific tag"
I'm using postgreSQL btw.

Answer 1

Here is a complete example that sets up relationships, creates some sample data, then performs your two queries.
Setup:
from datetime import datetime
from sqlalchemy import create_engine, Column, Integer, String, ForeignKey, not_
from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy.orm import sessionmaker, relationship
engine = create_engine('sqlite:///', echo=True)
Session = sessionmaker(bind=engine)
session = Session()
Base = declarative_base(bind=engine)
class User(Base):
__tablename__ = 'user'
id = Column(Integer, primary_key=True)
name = Column(String, nullable=False)
class Photo(Base):
__tablename__ = 'photo'
id = Column(Integer, primary_key=True)
name = Column(String, nullable=False)
user_id = Column(Integer, ForeignKey(User.id), nullable=False)
user = relationship(User, backref='photos')
class Tag(Base):
__tablename__ = 'tag'
id = Column(Integer, primary_key=True)
name = Column(String, nullable=False)
photo_id = Column(Integer, ForeignKey(Photo.id), nullable=False)
photo = relationship(Photo, backref='tags')
Base.metadata.create_all()
session.add(User(name='davidism', photos=[
Photo(name='sun', tags=[Tag(name='bright'), Tag(name='day')]),
Photo(name='moon', tags=[Tag(name='bright'), Tag(name='night')])
]))
session.add(User(name='eran', photos=[
Photo(name='party', tags=[Tag(name='people'), Tag(name='night')]),
Photo(name='cat')
]))
session.commit()
Query all photos with no tags at all:
no_tags = session.query(Photo).outerjoin(Photo.tags).filter(not_(Photo.tags.any())).all()
print 'no tags: ', len(no_tags)
Query all photos without the tag 'night':
not_night = session.query(Photo).outerjoin(Photo.tags).filter(not_(Photo.tags.any(Tag.name == 'night'))).all()
print 'not night: ', len(not_night)

Answer 2

Assuming existance of backrefs Tag.photo = relationship(Photo, backref='tags') and
Photo.user = relationship(User, backref="photos") both can be done using any construct. This might not generate the most optimal SQL SELECT statement, but it is a very clean sqlalchemy.
Part-1: "all the photos of the user with id "1234" that don't have a "cat" of version "2" tagged in them"
def get_user_photos_without_tag(user_id, tag_name, tag_version):
qry = (session.query(Photo)
.filter(~Photo.tags.any(and_(
Tag.tag_name == tag_name,
Tag.tag_version == tag_version))
)
.filter(Photo.user_id == user_id)
)
return qry.all()
photos = get_user_photos_without_tag(1234, 'cat', 2)
Part-2: "all the users who have at least one photo without a specific tag"
def get_user_with_photos_without_tag(tag_name, tag_version):
qry = (session.query(User)
.filter(User.photos.any(
~Photo.tags.any(and_(
Tag.tag_name == tag_name,
Tag.tag_version == tag_version))
))
)
return qry.all()
res = get_user_with_photos_without_tag('cat', 2)