我正在尝试学习炼金术中的关系,但是我很难让代码正常工作。首先,这个主题缺乏好的教程,但是我还是加入了一些零碎的信息来编写一些代码,但是我没有得到预期的结果
你能解释一下我在这段代码里做错了什么吗
from flask import Flask
from flask_sqlalchemy import SQLAlchemy
app = Flask(__name__)
app.config['SQLALCHEMY_DATABASE_URI'] = 'postgres://<my_user>:<my_password>@localhost/flask_apps'
db = SQLAlchemy(app)
class Universe(db.Model):
__tablename__ = 'universe'
id = db.Column(db.Integer, primary_key=True)
name = db.Column(db.String(30))
hero = db.relationship('Characters', backref='u_name', lazy='dynamic')
class Characters(db.Model):
__tablename__= 'characters'
id = db.Column(db.Integer, primary_key=True)
name = db.Column(db.String(50))
uni = db.Column(db.Integer, db.ForeignKey('universe.id'))
if __name__ == '__main__':
db.drop_all()
db.create_all()
dc = Universe(name='DC')
marvel = Universe(name='Marvel')
db.session.add(dc)
db.session.add(marvel)
db.session.commit()
flash = Characters(name='Flash', u_name=dc)
batman = Characters(name='Batman', u_name=dc)
arrow = Characters(name='Arrow', u_name=dc)
ironman = Characters(name='IronMan', u_name=marvel)
thor = Characters(name='Thor', u_name=marvel)
db.session.add(flash)
db.session.add(ironman)
db.session.commit()
u = Universe.query.filter_by(name='DC').first()
print(u.hero)
这就是我得到的结果
SELECT characters.id AS characters_id, characters.name AS characters_name, characters.uni AS characters_uni
FROM characters
WHERE %(param_1)s = characters.uni
Process finished with exit code 0
我希望能得到属于那个特定宇宙的人物的名字。我不太清楚我在这里遗漏了什么
我能够解决这个问题,问题是
hero
被误解为一个单值属性,而不是一个集合。所以使用下面的查询是可行的相关问题 更多 >
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