KeyError: 'security' in Flask_security?

Question

我正在使用 Flask 构建一个网站，我现在正尝试在其中使用 Flask_Security for token based authentication. I now want to get an auth_token from the user, for which I use the get_auth_token() method。不幸的是我得到了这条消息下面的堆栈跟踪。

有人知道怎么回事吗？欢迎所有提示！

Traceback (most recent call last):
  File "/usr/local/lib/python2.7/dist-packages/flask/app.py", line 1836, in __call__
    return self.wsgi_app(environ, start_response)
  File "/usr/local/lib/python2.7/dist-packages/flask/app.py", line 1820, in wsgi_app
    response = self.make_response(self.handle_exception(e))
  File "/usr/local/lib/python2.7/dist-packages/flask/app.py", line 1403, in handle_exception
    reraise(exc_type, exc_value, tb)
  File "/usr/local/lib/python2.7/dist-packages/flask/app.py", line 1817, in wsgi_app
    response = self.full_dispatch_request()
  File "/usr/local/lib/python2.7/dist-packages/flask/app.py", line 1477, in full_dispatch_request
    rv = self.handle_user_exception(e)
  File "/usr/local/lib/python2.7/dist-packages/flask/app.py", line 1381, in handle_user_exception
    reraise(exc_type, exc_value, tb)
  File "/usr/local/lib/python2.7/dist-packages/flask/app.py", line 1475, in full_dispatch_request
    rv = self.dispatch_request()
  File "/usr/local/lib/python2.7/dist-packages/flask/app.py", line 1461, in dispatch_request
    return self.view_functions[rule.endpoint](**req.view_args)
  File "/home/kramer65/myapp-kram/app/views/apiviews.py", line 33, in api_login
    return jsonify({'token': user.get_auth_token()})
  File "/usr/local/lib/python2.7/dist-packages/flask_security/core.py", line 313, in get_auth_token
    return _security.remember_token_serializer.dumps(data)
  File "/usr/local/lib/python2.7/dist-packages/werkzeug/local.py", line 338, in __getattr__
    return getattr(self._get_current_object(), name)
  File "/usr/local/lib/python2.7/dist-packages/werkzeug/local.py", line 297, in _get_current_object
    return self.__local()
  File "/usr/local/lib/python2.7/dist-packages/flask_security/core.py", line 30, in <lambda>
    _security = LocalProxy(lambda: current_app.extensions['security'])
KeyError: 'security'

Answer 1

您需要为您的应用程序初始化扩展程序。

security = Security()
security.init_app(app, user_datastore)

See the quickstart in the docs.

KeyError: 'security' in Flask_security?

KeyError: 'security' in Flask_security?

python

flask

peewee

flask-security