SQLAlchemy: filter by membership in at least one many-to-many related table

coredumperror picture coredumperror · Jun 25, 2011 · Viewed 19k times · Source

Using SQLAlchemy 0.7.1 and a MySQL 5.1 database, I've got a many-to-many relationship set up as follows:

user_groups = Table('user_groups', Base.metadata,
    Column('user_id', String(128), ForeignKey('users.username')),
    Column('group_id', Integer, ForeignKey('groups.id'))
)

class ZKUser(Base, ZKTableAudit):
    __tablename__ = 'users'

    username   = Column(String(128), primary_key=True)
    first_name = Column(String(512))
    last_name  = Column(String(512))

    groups = relationship(ZKGroup, secondary=user_groups, backref='users')

class ZKGroup(Base, ZKTableAudit):
    __tablename__ = 'groups'

    id          = Column(Integer, primary_key=True)
    name        = Column(String(512))

Users can belong to multiple Groups, and Groups can contain multiple Users.

What I'm trying to do is build a SQLAlchemy query that returns only the Users who belong to at least one Group out of a list of Groups.

I played around with the in_ function, but that only seems to work for testing scalar values for membership in a list. I'm not much of a SQL writer, so I don't even know what kind of SELECT statement this would require.

Answer

coredumperror picture coredumperror · Jun 28, 2011

OK, after a lot of research, I realized that it was my own ignorance of SQL terminology that was holding me back. My search for a solution to find users belonging to "at least one of" the list of groups should have been to find users belonging to "any" of the list of groups. The any ORM function from SQLAlchemy does exactly what I needed, like so:

session.query(ZKUser).filter(ZKUser.groups.any(ZKGroup.id.in_([1,2,3])))

That code emits this SQL (on MySQL 5.1):

SELECT * FROM users 
WHERE EXISTS (
    SELECT 1 FROM user_groups, groups 
    WHERE users.id = user_groups.contact_id 
        AND groups.id = user_groups.group_id 
        AND groups.id IN (%s, %s, %s)
    )