Flask-SQLALchemy：没有这样的表

Question

我试图让Flask-SQLAlchemy工作并遇到一些打嗝。 看看我正在使用的两个文件。 当我运行gwg.py和goto /util/db/create-all它会发出错误， no such table: stories 。 我以为我做的一切都是正确的; 有人可以指出我错过了什么或错了什么？ 它确实创建了data.db，但文件显示为0Kb

gwg.py：

application = Flask(__name__)
db = SQLAlchemy(application)

import models

# Utility
util = Blueprint('util', __name__, url_prefix='/util')

@util.route('/db/create-all/')
def db_create_all():
    db.create_all()
    return 'Tables created'

application.register_blueprint(util)

application.config['SQLALCHEMY_DATABASE_URI'] = 'sqlite:///data.db'
application.debug = True
application.run()

models.py：

from gwg import application, db

class Story(db.Model):
    __tablename__ = 'stories'
    id = db.Column(db.Integer, primary_key=True)
    name = db.Column(db.String(150))
    subscribed = db.Column(db.Boolean)

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

    def toggle_subscription(self):
        self.subscribed = False if self.subscribed else True

编辑

这是一个似乎触发错误的函数，但它不应该因为我特意先去/ util / db / create-all然后重新加载/。 即使在错误之后我转到/ util / db / create-all然后/仍然会得到相同的错误

@application.route('/')
def homepage():
    stories = models.Story.query.all()
    return render_template('index.html', stories=stories)

堆栈跟踪

sqlalchemy.exc.OperationalError
OperationalError: (OperationalError) no such table: stories u'SELECT stories.id AS stories_id, stories.name AS stories_name, stories.subscribed AS stories_subscribed \nFROM stories' ()

Traceback (most recent call last)
File "C:\Python27\lib\site-packages\flask\app.py", line 1836, in __call__
return self.wsgi_app(environ, start_response)
File "C:\Python27\lib\site-packages\flask\app.py", line 1820, in wsgi_app
response = self.make_response(self.handle_exception(e))
File "C:\Python27\lib\site-packages\flask\app.py", line 1403, in handle_exception
reraise(exc_type, exc_value, tb)
File "C:\Python27\lib\site-packages\flask\app.py", line 1817, in wsgi_app
response = self.full_dispatch_request()
File "C:\Python27\lib\site-packages\flask\app.py", line 1477, in full_dispatch_request
rv = self.handle_user_exception(e)
File "C:\Python27\lib\site-packages\flask\app.py", line 1381, in handle_user_exception
reraise(exc_type, exc_value, tb)
File "C:\Python27\lib\site-packages\flask\app.py", line 1475, in full_dispatch_request
rv = self.dispatch_request()
File "C:\Python27\lib\site-packages\flask\app.py", line 1461, in dispatch_request
return self.view_functions[rule.endpoint](**req.view_args)
File "C:\Users\kylee\Code\GWG\gwg.py", line 20, in homepage
stories = models.Story.query.all()
File "C:\Python27\lib\site-packages\sqlalchemy\orm\query.py", line 2104, in all
return list(self)
File "C:\Python27\lib\site-packages\sqlalchemy\orm\query.py", line 2216, in __iter__
return self._execute_and_instances(context)
File "C:\Python27\lib\site-packages\sqlalchemy\orm\query.py", line 2231, in _execute_and_instances
result = conn.execute(querycontext.statement, self._params)
File "C:\Python27\lib\site-packages\sqlalchemy\engine\base.py", line 662, in execute
params)
File "C:\Python27\lib\site-packages\sqlalchemy\engine\base.py", line 761, in _execute_clauseelement
compiled_sql, distilled_params
File "C:\Python27\lib\site-packages\sqlalchemy\engine\base.py", line 874, in _execute_context
context)
File "C:\Python27\lib\site-packages\sqlalchemy\engine\base.py", line 1024, in _handle_dbapi_exception
exc_info
File "C:\Python27\lib\site-packages\sqlalchemy\util\compat.py", line 163, in raise_from_cause
reraise(type(exception), exception, tb=exc_tb)
File "C:\Python27\lib\site-packages\sqlalchemy\engine\base.py", line 867, in _execute_context
context)
File "C:\Python27\lib\site-packages\sqlalchemy\engine\default.py", line 324, in do_execute
cursor.execute(statement, parameters)
OperationalError: (OperationalError) no such table: stories u'SELECT stories.id AS stories_id, stories.name AS stories_name, stories.subscribed AS stories_subscribed \nFROM stories' ()
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Answer 1

这里的问题是Python导入系统中的问题。 它可以简化为以下说明......

假设您在目录中有两个文件...

a.py：

print 'a.py is currently running as module {0}'.format(__name__)
import b
print 'a.py as module {0} is done'.format(__name__)

b.py：

print 'b.py is running in module {0}'.format(__name__)
import a
print 'b.py as module {0} is done'.format(__name__)

运行python a.py的结果如下......

a.py is currently running as module __main__
b.py is running in module b
a.py is currently running as module a
a.py as module a is done
b.py as module b is done
a.py as module __main__ is done

请注意，运行a.py时，模块名为__main__ 。

现在想想你拥有的代码。 在其中，您的a.py脚本会创建您的application和db对象。 但是，这些值不存储在名为a的模块中，它们存储在名为__main__ 。 因此， b.py尝试导入a ，它不会导入几秒钟前刚刚创建的值！ 相反，由于它没有找到模块a ，它创建一个NEW模块，运行a.py一个SECOND TIME并将结果存储到模块a 。

您可以使用我上面所示的打印语句来指导您完成整个过程，以确切了解正在发生的事情。 解决方案是确保您只调用a.py ONCE，并且当b.py导入a ，它将导入与a.py相同的值，而不是第二次导入它。

解决这个问题的最简单方法是创建一个仅用于运行应用程序的python脚本。 从gwg.py的末尾删除application.run() ，并添加以下脚本...

main.py：

from gwg import application
application.run()

然后使用python main.py运行。