python - SQLAlchemy count followers to a tag -

say have tag model:

class tag(db.model):     id = db.column(db.integer, primary_key = true)     name =  db.column(db.string(240))      def followerranked(self):                 #return query(followers, func.count(followers.c.tag_id)).outerjoin(tag)??         #return tag.query.join(followers, (followers.c.tag_id == self.id).count()??         #i not sure how 

say have subscriptions table relationship.

followers = db.table('followers',     db.column('follower_id', db.integer, db.foreignkey('user.id')),     db.column('tag_id', db.integer, db.foreignkey('tag.id')) ) 

i want loop through tags... in desc order number of followers per tag. think need group_by somewhere.

i want able loop through:

tags.followersranked().items

how do this?

edit:

apparently, becomes difficult in model, i'm trying in view instead:

tags = db.session.query(models.tag, \ func.count(models.followers.c.follower_id).label('total'), models.tag.name, \ models.tag.title).join(models.followers).group_by(models.tag).order_by('total desc') p = pagination(tags, page, posts_per_page, tags.count(), tags) 

the issue is, doesn't show tags 0 subscribers.

change join() outerjoin() , use coalesce() change nulls 0s:

tags = db.session.query(     db.coalesce(db.func.count(models.followers.c.follower_id), 0).label('total'),     models.tag.name,     models.tag.title ).outerjoin(models.followers).group_by(     models.tag.name,     models.tag.title ).order_by('total desc') 

that should trick.