如何在SQLAlchemy中按多对多关系进行排序? [英] How to order by count of many-to-many relationship in SQLAlchemy?
问题描述
我想按表上多对多关系的数量进行排序.
I'd like to order by the count of a many-to-many relationship on a table.
我的出发点是对此线程.但是，在我的情况下，涉及到第三个表，我需要我的backref被加入"而不是动态".
My starting point was the (excellent) response to this thread. However, in my case, there is a third table involved and I need my backrefs to be 'joined' rather than 'dynamic'.
当我尝试解决该问题时，这两个更改的组合会导致ProgrammingErrors.
The combination of those two changes results in ProgrammingErrors when I try the solution to that question.
这是我的设置.我和用户之间有相同的多对多关系.我在帖子和标题之间也有一对多的关系.我想要的是按喜欢每个帖子的用户数量排序的所有帖子的列表.
Here's my setup. I have the same many-to-many relationship between User and Post. I also have a one-to-many relationship between Post and Heading. What I'd like is the list of all posts ordered by the number of users that like each post.
from flask import Flask
from flask.ext.sqlalchemy import SQLAlchemy
from sqlalchemy.orm import relationship, backref
from sqlalchemy import func
app = Flask(__name__)
app.config['SQLALCHEMY_DATABASE_URI'] = 'postgres://xxx'
db = SQLAlchemy(app)
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='joined'))
def __repr__(self):
return "<Post('%s')>" % self.title
class Heading(db.Model):
id = db.Column(db.Integer, primary_key=True)
name = db.Column(db.String(255))
post_id = db.Column(db.Integer, db.ForeignKey('post.id'))
post = relationship('Post', backref=backref('headings', lazy='joined'))
def __repr__(self):
return "<Category('%s')>" % self.name
# Three users
user1 = User(username='user1')
user2 = User(username='user2')
user3 = User(username='user3')
# Two posts
post1 = Post(title='post1')
heading1 = Heading(name='heading1')
heading2 = Heading(name='heading2')
post1.headings = [heading1, heading2]
post1.likes = [user1, user2]
post2 = Post(title='post2')
heading3 = Heading(name='heading3')
post2.headings = [heading3]
post2.likes = [user3]
db.create_all()
db.session.add_all([user1, user2, user3, post1, post2, heading1, heading2, heading3])
db.session.commit()
现在，如果我在另一个问题中尝试解决方案:
Now if I try the solution in the other question:
In [3]: db.session.query(Post, func.count(likes.c.user_id).label('total')).join(likes).group_by(Post).order_by('total DESC').all()
---------------------------------------------------------------------------
ProgrammingError Traceback (most recent call last)
<ipython-input-3-b804fb4b0893> in <module>()
----> 1 db.session.query(Post, func.count(likes.c.user_id).label('total')).join(likes).group_by(Post).order_by('total DESC').all()
ProgrammingError: (ProgrammingError) column "heading_1.id" must appear in the GROUP BY clause or be used in an aggregate function
LINE 1: ...tle AS post_title, count(likes.user_id) AS total, heading_1....
^
'SELECT post.id AS post_id, post.title AS post_title, count(likes.user_id) AS total, heading_1.id AS heading_1_id, heading_1.name AS heading_1_name, heading_1.post_id AS heading_1_post_id \nFROM post JOIN likes ON post.id = likes.post_id LEFT OUTER JOIN heading AS heading_1 ON post.id = heading_1.post_id GROUP BY post.id, post.title ORDER BY total DESC' {}
如果我尝试将Heading添加到我的group_by中，那么天真地(并且在SQL和SQLAlchemy方面我很天真).
Naively (and I'm pretty naive when it comes to SQL and SQLAlchemy), if I try to add Heading to my group_by:
In [6]: db.session.query(Post, func.count(likes.c.user_id).label('total')).join(likes).group_by(Post, Heading).order_by('total DESC').all()
---------------------------------------------------------------------------
ProgrammingError Traceback (most recent call last)
<ipython-input-6-8efd6a8314f8> in <module>()
----> 1 db.session.query(Post, func.count(likes.c.user_id).label('total')).join(likes).group_by(Post, Heading).order_by('total DESC').all()
ProgrammingError: (ProgrammingError) invalid reference to FROM-clause entry for table "heading"
LINE 2: ...= heading_1.post_id GROUP BY post.id, post.title, heading.id...
^
HINT: Perhaps you meant to reference the table alias "heading_1".
'SELECT post.id AS post_id, post.title AS post_title, count(likes.user_id) AS total, heading_1.id AS heading_1_id, heading_1.name AS heading_1_name, heading_1.post_id AS heading_1_post_id \nFROM post JOIN likes ON post.id = likes.post_id LEFT OUTER JOIN heading AS heading_1 ON post.id = heading_1.post_id GROUP BY post.id, post.title, heading.id, heading.name, heading.post_id ORDER BY total DESC' {}
如何获取所需的查询?我可能正在做一些非常愚蠢的事情.
How can I get my desired query to work? I'm probably doing something really dumb.
推荐答案
正如您所提到的，提到的解决方案不起作用的真正原因是标题的joinedload.您可以为此做些事情:
As you mentioned, the real reason for the mentioned solution not working is the joinedload of the Headings. Few things you can do about it:
q = (db.session.query(Post, func.count(likes.c.user_id).label("total"))
.options(lazyload(Post.headings)) # disable joined-load
.join(likes) .group_by(Post) .order_by('total DESC')
)
如果确实需要标题，则可以使用subqueryload代替:
In case you really need the headings, you can use subqueryload instead:
q = (db.session.query(Post, func.count(likes.c.user_id).label("total"))
.options(subqueryload(Post.headings))
.join(likes) .group_by(Post) .order_by('total DESC')
)
选项2:使用子查询
subq = (db.session.query( Post.id.label("post_id"),
func.count(likes.c.user_id).label("num_likes"))
.outerjoin(likes).group_by(Post.id)
).subquery("subq")
q = (db.session.query(Post, subq.c.num_likes)
.join(subq, Post.id == subq.c.post_id)
.group_by(Post) .order_by(subq.c.num_likes.desc())
)
选项3:使用混合属性
它可能不是最有效的，但是可以保持干净.
Option-3: use Hybrid Attributes
It might not be the most efficient, but keeps things clean.
在Post模型中添加以下内容:
Add following to the Post model:
from sqlalchemy.ext.hybrid import hybrid_property
class Post(db.Model):
# ...
@hybrid_property
def num_likes(self):
return len(self.likes)
@num_likes.expression
def _num_likes_expression(cls):
return (db.select([db.func.count(likes.c.user_id).label("num_likes")])
.where(likes.c.post_id == cls.id)
.label("total_likes")
)
稍后，您可以采用一种简洁的方式来组成查询:
Later you can compose a query in an clean way:
q = db.session.query(Post, Post.num_likes).order_by(Post.num_likes.desc())
# use this if you need it data sorted, but do not care how many likes are there
#q = db.session.query(Post).order_by(Post.num_likes.desc())
这篇关于如何在SQLAlchemy中按多对多关系进行排序?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
