'没有找到申请.要么在视图函数内部工作,要么推送应用程序上下文.

Ome*_*ick 27 python flask flask-sqlalchemy

我正在尝试将Flask-SQLAlchemy模型分成单独的文件.当我试着奔跑时,db.create_all()我得到了No application found. Either work inside a view function or push an application context.

shared/db.py:

from flask_sqlalchemy import SQLAlchemy

db = SQLAlchemy()
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app.py:

from flask import Flask
from flask_sqlalchemy import SQLAlchemy
from shared.db import db

app = Flask(__name__)
app.config['SQLALCHEMY_DATABASE_URI'] = 'My connection string'
db.init_app(app)
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user.py:

from shared.db import db

class User(db.Model):
    id = db.Column(db.Integer, primary_key=True)
    email_address = db.Column(db.String(300), unique=True, nullable=False)
    password = db.Column(db.Text, nullable=False)
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Sht*_*zut 49

使用with app.app_context()推的应用程序上下文该块.

app = Flask(__name__)
app.config['SQLALCHEMY_DATABASE_URI'] = 'My connection string'
db.init_app(app)

with app.app_context():
    db.create_all()
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  • 这对我在 app = create_app() app.app_context().push() 之后有所帮助,我尝试根据 SqlAlchemy 从控制台执行此操作,它无法自动将您的数据库绑定到应用程序 - 此处有更多信息:https:// Flask-sqlalchemy.palletsprojects.com/en/2.x/contexts/ (3认同)
  • 我认为你也需要创建上下文 - 如果我没记错的话 - >>> from yourapp import create_app >>> app = create_app() >>> app.app_context().push() (3认同)