标签：peewee python flask flask-security

我正在使用Flask建立一个网站,我现在正在尝试使用Flask_Security进行基于令牌的身份验证.我现在想从用户那里获得一个auth_token,我使用了get_auth_token() method.不幸的是我在这条消息下面得到了栈跟踪.

有人知道什么是错的吗？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'

解决方法:

您需要初始化应用的扩展程序.

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

See the quickstart in the docs.

标签：peewee,python,flask,flask-security
来源： https://codeday.me/bug/20190830/1765508.html

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