我正在使用Flask和SQLAlchemy。我已经使用了我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,我得到了以下错误:
>>> 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.
我该怎么解决这个问题?
- 代码:**
- 管理. 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()
- 初始化. 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
- 数据库. py**:
from flask.ext.sqlalchemy import SQLAlchemy
db = SQLAlchemy()
- 型号. 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
- 控制器. 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")
3条答案
按热度按时间wj8zmpe11#
尝试添加
到
__tablename__=
下的用户类干杯
pzfprimi2#
另一种选择是,如果你已经存在的数据库中没有数据,那么你可以删除它,重新创建一个没有表的数据库,然后交互式地运行你的“models.py“脚本(你需要在底部添加一些代码来实现这一点),然后在交互式Python控制台中执行“db.create_all()”,它应该会基于你的类创建表。
xmakbtuz3#
我有同样的问题,并在这里找到解决方案:https://github.com/pallets-eco/flask-sqlalchemy/issues/672#issuecomment-478195961
1.在磁盘上留下.pyc文件。