Python 炼金术-顺序显示的追随者人数
我正在尝试创建一个查询,该查询将返回数据库中按收藏该节目的用户数排序的所有节目 简化的工作代码:Python 炼金术-顺序显示的追随者人数,python,sqlalchemy,flask,Python,Sqlalchemy,Flask,我正在尝试创建一个查询,该查询将返回数据库中按收藏该节目的用户数排序的所有节目 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('
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()
但这在SQL语法中引发了错误,我试图搜索一些东西,但找不到任何有效的东西
任何帮助都将不胜感激。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()
这可能是它的重复,因为我使用了很多,我使用了交叉表,并且这个例子有外键。最终的工作解决方案:
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).外套(sub,Serie.id==sub.c.series\u id).订购人(db.desc('count')).all()@OndrejFabry请将您的解决方案添加为已接受的答案,Ross.谢谢提醒!
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()