9

我正在尝试让 Flask-SQLAlchemy 工作并遇到一些问题。看看我正在使用的两个文件。当我运行gwg.py并转到/util/db/create-all它时，它会吐出一个错误no such table: stories。我以为我做的一切都是正确的；有人可以指出我缺少什么或有什么问题吗？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()

模型.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

这是似乎触发错误的函数，但它不应该因为我专门先去 /util/db/create-all 然后重新加载 /。即使在错误之后我去 /util/db/create-all 然后 / 仍然得到同样的错误

@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' ()
The debugger caught an exception in your WSGI application. You can now look at the traceback which led to the error.
To switch between the interactive traceback and the plaintext one, you can click on the "Traceback" headline. From the text traceback you can also create a paste of it. For code execution mouse-over the frame you want to debug and click on the console icon on the right side.

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4

1 回答 1

13

这里的问题是 Python 导入系统中的一个问题。可以简化为以下解释...

假设您在一个目录中有两个文件...

一个.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__)

运行结果python a.py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，当a.py运行时，该模块被调用__main__。

现在想想你拥有的代码。在其中，您的a.py脚本创建您的application和db对象。但是，这些值并未存储在名为 的模块中a，而是存储在名为 的模块中__main__。因此，b.py尝试导入a，它不是导入您几秒钟前刚刚创建的值！相反，由于它没有找到模块a，它创建一个新模块，运行a.py第二次并将结果存储到模块a中。

您可以使用我上面显示的打印语句，希望能引导您完成整个过程，确切了解发生了什么。解决方案是确保您只调用a.py一次，并且在b.py导入a时它将导入相同的值a.py而不是第二次导入它。

解决此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...

主要.py：

from gwg import application
application.run()

然后使用python main.py.

于 2013-07-03T16:10:05.507 回答