我正在研究烧瓶关系教程。我正在尝试建立一个数据库，用于存储小狗，有关其玩具的信息以及有关其所有者的信息， basic.py：
# basic.py
import os
from flask import Flask
from flask_sqlalchemy import SQLAlchemy
from flask_migrate import Migrate
basedir = os.path.abspath(os.path.dirname(__file__))
app = Flask(__name__)
app.config['SQLALCHEMY_DATABASE_URI'] = 'sqlite:///'+os.path.join(basedir, 'data.sqlite')
app.config['SQLALCHEMY_TRACK_MODIFICATIONS'] = False
db = SQLAlchemy(app)
Migrate(app, db)
class Puppy(db.Model):
__tablename__ = 'puppies'
id = db.Column(db.Integer, primary_key = True)
name = db.Column(db.Text)
toys = db.relationship('Toy', backref = 'puppy', lazy = 'dynamic')
owner = db.relationship('Owner', backref = 'puppy', uselist = False)
def __init__(self, name):
self.name = name
def __repr__(self):
if self.owner:
return "Puppy name is {self.name} and owner is {self.owner}"
else:
return "Puppy is {self.name} and has no owner yet"
def report_toys(self):
print("Here are my toys:")
for toy in self.toys:
print(toy.item_name)
class Toy(db.Model):
__tablename__ = 'toys'
id = db.Column(db.Integer, primary_key = True)
item_name = db.Column(db.Text)
puppy_id = db.Column(db.Integer, db.ForeignKey('puppies.id'))
def __init__(self, item_name, puppy_id):
self.item_name = item_name
self.puppy_id = puppy_id
class Owner(db.Model):
__tablename__ = 'owners'
id = db.Column(db.Integer, primary_key = True)
name = db.Column(db.Text)
puppy_id = db.Column(db.Integer, db.ForeignKey('puppies.id'))
def __init__(self, name, puppy_id):
self.name = name
self.puppy_id = puppy_id
然后我在models.py中建立一个手动数据库：
# models.py
from basic import db, Puppy, Owner, Toy
rufus = Puppy('Rufus')
fido = Puppy('Fido')
db.session.add_all([rufus, fido])
db.session.commit
rufus = Puppy.query.filter_by(name = 'Rufus').first()
kate = Owner('Kate', rufus.id)
toy1 = Toy('Chew Toy', rufus.id)
toy2 = Toy('Ball', rufus.id)
db.session.add_all([kate, toy1, toy2])
db.session.commit()
但是我遇到一个错误：
cursor.execute(statement, parameters)
sqlalchemy.exc.OperationalError: (raised as a result of Query-invoked autoflush; consider using a session.no_autoflush block if this flush is occurring prematurely)
(sqlite3.OperationalError) no such table: puppies
[SQL: INSERT INTO puppies (name) VALUES (?)]
[parameters: ('Rufus',)]
(Background on this error at: http://sqlalche.me/e/13/e3q8)
我是新来的，因此不胜感激。
答案 0 :(得分：2)
您创建数据库了吗？
您需要导入db对象和模型，然后创建数据库，如下所示：
(venv) $ flask shell
>>> from basic import db
>>> from basic import Puppy, Toy, Owner
>>> db.create_all()
让我知道这是否可以解决您的问题
答案 1 :(得分：2)
错误消息显示数据库中没有幼犬表。要将迁移应用于数据库，在使用flask-migrate时需要一些步骤。
答案 2 :(得分：0)
由于使用的是flask_migrate，因此需要打开一个终端，请确保位于项目的同一工作目录中并运行以下命令：
您需要先运行它，然后再运行实际的Flask应用程序
在环境文件夹中的
- CMD：
export class HeroFormComponent implements OnInit {
@ViewChild("heroForm", null) heroForm: NgForm;
queryParams={};
constructor(private route: ActivatedRoute) {}
ngOnInit() {
this.queryParams = { fname: "aaa", lname: "bbb" };
}
}
通常，如果您不使用flask_migrate，而仅使用flask_sqlalchemy或SQLAlchemy vanilla，则应运行此程序。
>>> set FLASK_APP= app.py (optional)
>>> flask db init
>>> flask db migrate -m 'Any name you want' ---> what's inside the '' is the name of migrati
>>> flask db upgrade (perform the actual migration)
编辑
每次您都从实例的数据库表中添加（或删除）某些东西
>>> from app import db
>>> db.create_all()
>>> exit() #or ctrl + z then enter
成为--->
class Puppy(db.Model):
__tablename__ = 'puppies'
id = db.Column(db.Integer, primary_key = True)
name = db.Column(db.Text)
toys = db.relationship('Toy', backref = 'puppy', lazy = 'dynamic')
owner = db.relationship('Owner', backref = 'puppy', uselist = False)
然后，您应该删除数据库文件（在您的情况下为“ data.sqlite”）（在工作目录中），然后再次运行上述代码，