SQLAlchemy继承不起作用

10 python postgresql flask flask-sqlalchemy

我正在使用Flask和SQLAlchemy.我使用了自己的抽象基类和继承.当我尝试在python shell中使用我的模型时,我收到以下错误:

>>> from schedule.models import Task
Traceback (most recent call last):
  File "<console>", line 1, in <module>
  File "/home/teelf/projects/schedule/server/schedule/models.py", line 14, in <module>
    class User(Base):
  File "/home/teelf/projects/schedule/server/venv/lib/python3.4/site-packages/flask_sqlalchemy/__init__.py", line 536, in __init__
    DeclarativeMeta.__init__(self, name, bases, d)
  File "/home/teelf/projects/schedule/server/venv/lib/python3.4/site-packages/sqlalchemy/ext/declarative/api.py", line 55, in __init__
    _as_declarative(cls, classname, cls.__dict__)
  File "/home/teelf/projects/schedule/server/venv/lib/python3.4/site-packages/sqlalchemy/ext/declarative/base.py", line 254, in _as_declarative
    **table_kw)
  File "/home/teelf/projects/schedule/server/venv/lib/python3.4/site-packages/sqlalchemy/sql/schema.py", line 393, in __new__
    "existing Table object." % key)
sqlalchemy.exc.InvalidRequestError: Table 'user' is already defined for this MetaData instance.  Specify 'extend_existing=True' to redefine options and columns
 on an existing Table object.
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我该如何解决?

码:

manage.py:

#!/usr/bin/env python

import os, sys

sys.path.append(os.path.abspath(os.path.join(os.path.dirname(__file__), '..')))

from flask.ext.migrate import Migrate, MigrateCommand
from flask.ext.script import Manager

from server import create_app
from database import db


app = create_app("config")

migrate = Migrate(app, db)
manager = Manager(app)

manager.add_command("db", MigrateCommand)

if __name__ == "__main__":
    manager.run()
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__init__.py:

from flask import Flask

from flask.ext.login import LoginManager

from database import db
from api import api
from server.schedule.controllers import mod_schedule


def create_app(config):
    # initialize Flask
    app = Flask(__name__)

    # load configuration file
    app.config.from_object(config)

    # initialize database
    db.init_app(app)

    api.init_app(app)

    # initialize flask-login
    login_manager = LoginManager(app)

    # register blueprints
    app.register_blueprint(mod_schedule)

    return app
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database.py:

from flask.ext.sqlalchemy import SQLAlchemy


db = SQLAlchemy()
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models.py:

from sqlalchemy.dialects.postgresql import UUID

from database import db


class Base(db.Model):
    __abstract__ = True

    id = db.Column(UUID, primary_key=True)


class User(Base):
    __tablename__ = "user"

    username = db.Column(db.String)
    password = db.Column(db.String)
    first_name = db.Column(db.String)
    last_name = db.Column(db.String)
    authenticated = db.Column(db.Boolean, default=False)

    def __init__(self, first_name, last_name, username):
        self.first_name = first_name
        self.last_name = last_name
        self.username = username

    def is_active(self):
        """ All users are active """
        return True

    def get_id(self):
        return self.username

    def is_authenticated(self):
        return self.authenticated

    def is_anonymous(self):
        """ Anonymous users are not supported"""
        return False
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controllers.py:

from flask import Blueprint

from flask.ext.restful import reqparse, Resource

from api import api
from server.schedule.models import User


mod_schedule = Blueprint("schedule",  __name__, url_prefix="/schedule")


class Task(Resource):
    def put(self):
        pass

    def get(self):
        pass

    def delete(self):
        pass


api.add_resource(Task, "/tasks/<int:id>", endpoint="task")
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Jef*_* R. 24

尝试添加

 __table_args__ = {'extend_existing': True} 
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到你的用户类 __tablename__=

干杯

  • 虽然这在某些情况下很有用,但通常不需要这样做,并且表示多次创建表定义 - 通常是偶然的.把它作为通用"只是复制并粘贴到你的代码"解决方案是误导性的,只是隐藏真正的错误. (10认同)
  • 你是从哪里学到这一点的,你能分享一下SQLalchemy文档中的案例吗?(你回答的正是我需要的) (5认同)
  • 这是 sqlalchemy 文档的[链接](http://docs.sqlalchemy.org/en/latest/orm/extensions/declarative/table_config.html?highlight=__table_args__),其中提到了 \__\_table_args\_\_ dict 。 (3认同)