python - Flask-SQLALchemy:没有这样的表

Question

我正在尝试让 Flask-SQLAlchemy 工作并遇到一些问题。看一下我正在使用的两个文件。当我运行 gwg.py 并转到 /util/db/create-all 时，它吐出错误 no such table: stories。我以为我做的一切都是正确的；有人可以指出我遗漏了什么或出了什么问题吗？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

这里是似乎触发错误的函数，但它不应该，因为我特别先转到/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 导入系统中的一个问题。可以简化为下面的解释...

假设您在一个目录中有两个文件...

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，当 a.py 运行时，该模块被称为 __main__。

现在想想你的代码。在其中，您的 a.py 脚本会创建您的 application 和 db 对象。但是，这些值并未存储在名为 a 的模块中，它们存储在名为 __main__ 的模块中。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，它创建了一个新模块，第二次运行 a.py 并将结果存储到模块 a.

您可以使用我上面显示的打印语句来指导您完成整个过程，以准确了解发生了什么。解决方案是确保您只调用一次 a.py，当 b.py 导入 a 时，它将导入相同的值作为 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 运行。