SQLAlchemy ordering by count on a many to many relationship
This is a simplified example of my current models (I'm using the Flask SQLAlchemy extension):
like = db.Table(
'like',
db.Column('uid', db.Integer, db.ForeignKey('users.id')),
db.Column('pid', db.Integer, db.ForeignKey('posts.id'开发者_JAVA百科))
)
class User(db.Model):
__tablename__ = 'users'
id = db.Column(db.Integer, primary_key = True)
username = db.Column(db.String(20))
class Post(db.Model):
__tablename__ = 'posts'
id = db.Column(db.Integer, primary_key = True)
title = db.Column(db.String(255))
likes = db.relationship(
'User',
secondary = like,
backref = db.backref('likes', lazy = 'dynamic'),
lazy = 'dynamic'
)
I'm trying to order Post
's by the amount of likes it has.
This is the query I'm basically trying to issue:
SELECT p.*, COUNT(l.`pid`) as `likes`
FROM `posts` as p
LEFT JOIN `like` as l
ON p.`id` = l.`pid`
GROUP BY p.`id`
ORDER BY `likes` DESC
I just haven't been able to get anything working on the SQLAlchemy side of things.
Thanks for any help anyone can offer.
I haven't used SQLAlchemy much so I figured I'd give it a shot. I didn't try to use your models, I just wrote some new ones (similar enough though):
likes = db.Table('likes',
db.Column('user_id', db.Integer, db.ForeignKey('user.id')),
db.Column('post_id', db.Integer, db.ForeignKey('post.id'))
)
class User(db.Model):
id = db.Column(db.Integer, primary_key=True)
username = db.Column(db.String(20))
def __repr__(self):
return "<User('%s')>" % self.username
class Post(db.Model):
id = db.Column(db.Integer, primary_key=True)
title = db.Column(db.String(255))
likes = db.relationship('User', secondary = likes,
backref = db.backref('posts', lazy='dynamic'))
def __repr__(self):
return "<Post('%s')>" % self.title
You want to join the likes
table, use func.count
to count likes, group_by
Post
and then use order_by
:
db.session.query(Post, func.count(likes.c.user_id).label('total'))\
.join(likes).group_by(Post).order_by('total DESC')
I found the ORM tutorial and the rest of the SQLAlchemy documentation very useful.
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