flask_migrate KeyError：“迁移”

Question

我目前正在学习烧瓶，我正在研究数据库之间的关系，但是我正在尝试cmd中的以下命令：

set FLASK_APP=app4.py
flask db init

当我运行它时，我得到以下信息：

Traceback (most recent call last):
  File "c:\users\admin\appdata\local\programs\python\python37-32\lib\runpy.py", line 193, in _run_module_as_main
    "__main__", mod_spec)
  File "c:\users\admin\appdata\local\programs\python\python37-32\lib\runpy.py", line 85, in _run_code
    exec(code, run_globals)
  File "C:\Users\admin\AppData\Local\Programs\Python\Python37-32\Scripts\flask.exe\__main__.py", line 9, in <module>
  File "c:\users\admin\appdata\local\programs\python\python37-32\lib\site-packages\flask\cli.py", line 966, in main
    cli.main(prog_name="python -m flask" if as_module else None)
  File "c:\users\admin\appdata\local\programs\python\python37-32\lib\site-packages\flask\cli.py", line 586, in main
    return super(FlaskGroup, self).main(*args, **kwargs)
  File "c:\users\admin\appdata\local\programs\python\python37-32\lib\site-packages\click\core.py", line 717, in main
    rv = self.invoke(ctx)
  File "c:\users\admin\appdata\local\programs\python\python37-32\lib\site-packages\click\core.py", line 1137, in invoke
    return _process_result(sub_ctx.command.invoke(sub_ctx))
  File "c:\users\admin\appdata\local\programs\python\python37-32\lib\site-packages\click\core.py", line 1137, in invoke
    return _process_result(sub_ctx.command.invoke(sub_ctx))
  File "c:\users\admin\appdata\local\programs\python\python37-32\lib\site-packages\click\core.py", line 956, in invoke
    return ctx.invoke(self.callback, **ctx.params)
  File "c:\users\admin\appdata\local\programs\python\python37-32\lib\site-packages\click\core.py", line 555, in invoke
    return callback(*args, **kwargs)
  File "c:\users\admin\appdata\local\programs\python\python37-32\lib\site-packages\click\decorators.py", line 17, in new_func
    return f(get_current_context(), *args, **kwargs)
  File "c:\users\admin\appdata\local\programs\python\python37-32\lib\site-packages\flask\cli.py", line 426, in decorator
    return __ctx.invoke(f, *args, **kwargs)
  File "c:\users\admin\appdata\local\programs\python\python37-32\lib\site-packages\click\core.py", line 555, in invoke
    return callback(*args, **kwargs)
  File "c:\users\admin\appdata\local\programs\python\python37-32\lib\site-packages\flask_migrate\cli.py", line 31, in init
    _init(directory, multidb)
  File "c:\users\admin\appdata\local\programs\python\python37-32\lib\site-packages\flask_migrate\__init__.py", line 96, in wrapped
    f(*args, **kwargs)
  File "c:\users\admin\appdata\local\programs\python\python37-32\lib\site-packages\flask_migrate\__init__.py", line 126, in init
    directory = current_app.extensions['migrate'].directory
KeyError: 'migrate'

我真的不知道我做错了什么，任何帮助都是非常感激的。下面是我目前拥有的python脚本：

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 = Migrate()
migrate.init_app(app, db)



class Puppies(db.Model):

    __tablename__ = 'Puppies'

    id = db.Column(db.Integer, primary_key=True)
    name = db.Column(db.Text)
    toys = db.relationship('Toy', backref='Puppies', lazy='dynamic') #Connects to the Toy model (the class below) | connects the puppy to many toys | will return list of toys
    owner = db.relationship('Owner', backref='Puppies', uselist=False) #uselist=False will ensure it doesn't bring a list of items, it will return 1.

    def __init__(self,name):
        self.name = name

    def __repr__(self):
        if self.owner:
            return f"Puppy Name: {self.name} | Owner: {self.owner.name}"
        else:
            return f"Puppy Name: {self.name} | The puppy currently has no owner."

    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)
    puppies_id = db.Column(db.Integer, db.ForeignKey(Puppies.id)) #this will get the id from the Puppies table (the above class)

    def __init__(self, item_name, puppies_id):
        self.item_name = item_name
        self.puppies_id = puppies_id


class Owner(db.Model):

    __tablename__ = 'Owners'

    id = db.Column(db.Integer, primary_key=True)
    name = db.Column(db.Text)
    puppies_id = db.Column(db.Integer, db.ForeignKey(Puppies.id)) #this will get the id from the Puppies table

    def __init__(self, name, puppies_id):
        self.name = name
        self.puppies_id = puppies_id

Answer 1

这个错误显然是一个烧瓶分贝。希望您已经导入了flask-migrate和flask-sqlalchemy，在__init__.py中注册flask-migrate的方式是不正确的。

案例1:

如果在应用程序中使用应用程序工厂，则需要使用migrate.init_app(app, db)。所以您的__init__.py看起来是这样的：

from flask_sqlalchemy import SQLAlchemy
from flask_migrate import Migrate
app.config.from_object(Config)

db = SQLAlchemy()
migrate = Migrate()

def create_app(config_class=Config):
    app = Flask(__name__)
    app.config.from_object(config_class)

    db.init_app(app)
    migrate.init_app(app, db)
    # ...

上面，您的工厂功能是create_app()。

案例2:

如果您没有像在脚本中那样使用工厂函数，那么就不要使用migrate.init_app(app, db)。相反，让您的__init__.py足够简单：

from flask_sqlalchemy import SQLAlchemy
from flask_migrate import Migrate
app.config.from_object(Config)

db = SQLAlchemy(app)
migrate = Migrate(app, db)

# ...

使用这两种方法，您都可以运行迁移：

$ flask db init
$ flask db migrate -m '<your table>'