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我正在尝试创建一个查询,该查询将返回数据库中的所有节目,这些节目按收藏它的用户数量排序。

简化的工作代码:

from flask import Flask
from flask.ext.sqlalchemy import SQLAlchemy
from sqlalchemy.sql import func
import logging

app = Flask(__name__)
db = SQLAlchemy(app)

logging.basicConfig()
logging.getLogger('sqlalchemy.engine').setLevel(logging.INFO)

favorite_series = db.Table('favorite_series',
        db.Column('user_id', db.Integer, db.ForeignKey('users.id')),
        db.Column('series_id', db.Integer, db.ForeignKey('series.id')))

class User(db.Model):
    __tablename__ = 'users'

    id = db.Column(db.Integer, primary_key=True)
    name = db.Column(db.String)
    favorite_series = db.relationship('Serie', secondary=favorite_series,
            backref=db.backref('users', lazy='dynamic'))

    def __repr__(self):
        return '<User {0}>'.format(self.name)


class Serie(db.Model):
    __tablename__ = 'series'

    id = db.Column(db.Integer, primary_key=True)
    name = db.Column(db.String)

    def __repr__(self):
        return '<Serie {0}>'.format(self.name)

u1 = User()
u1.name = 'user1'

u2 = User()
u2.name = 'user2'

u3 = User()
u3.name = 'user3'

s1 = Serie()
s1.name = 'Serie1'

s2 = Serie()
s2.name = 'Serie2'

s3 = Serie()
s3.name = 'Serie3'

s4 = Serie()
s4.name = 'Serie4'

s5 = Serie()
s5.name = 'Serie5'

u1.favorite_series.extend([s1, s3, s5])
u2.favorite_series.extend([s1, ])
u3.favorite_series.extend([s1, s2, s3])

u1.favorite_series.extend([s1, s2])
db.session.add(u1)
db.session.add(u2)
db.session.add(u3)
db.session.add(s1)
db.session.add(s2)
db.session.add(s3)
db.session.add(s4)
db.session.add(s5)
db.create_all()
db.session.commit()

我尝试通过以下方式检索它们:

shows = Serie.query.join(Serie.users).order_by(func.count(Serie.users)).all()

print shows

但这会在 SQL 语法中引发错误,我试图搜索一些东西但无法找到任何有用的东西。

任何帮助,将不胜感激。

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1 回答 1

2

工作解决方案:

sub = db.session.query(favorite_series.c.series_id, func.count(favorite_series.c.user_id).label('count')).group_by(favorite_series.c‌​.series_id).subquery()
shows = db.session.query(Serie, sub.c.count).outerjoin(sub, Serie.id == sub.c.series_id).order_by(db.desc('count')).all()
于 2012-11-21T11:15:27.507 回答