我想通过桌子上的多对多关系来计算。
我的出发点是对this线程的（优秀）响应。但是，在我的情况下，涉及第三个表，我需要我的backrefs'加入'而不是'动态'。
当我尝试解决该问题的解决方案时，这两个更改的组合会导致编程错误。
这是我的设置。我在User和Post之间拥有相同的多对多关系。 Post和Heading之间也有一对多的关系。我想要的是按照每个帖子的用户数量排序的所有帖子的列表。
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：
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' {}
Naively（对于SQL和SQLAlchemy来说，我很天真），如果我尝试将Heading添加到我的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' {}
如何让我想要的查询工作？我可能正在做一些非常愚蠢的事情。
答案 0 :(得分：3)
正如您所提到的，上述解决方案不起作用的真正原因是标题的joinedload。你可以做的很少事情：
joinedload 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')
)
如果您确实需要标题，可以改为使用subqueryload：
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')
)
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())
)
它可能不是最有效的，但保持清洁。
将以下内容添加到Post模型中：
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")
)
稍后您可以以干净的方式撰写查询：
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())