def open_resource(self, resource, mode='rb'):
        """Opens a resource from the application's resource folder.  To see
        how this works, consider the following folder structure::

            /myapplication.py
            /schema.sql
            /static
                /style.css
            /templates
                /layout.html
                /index.html

        If you want to open the `schema.sql` file you would do the
        following::

            with app.open_resource('schema.sql') as f:
                contents = f.read()
                do_something_with(contents)

        :param resource: the name of the resource.  To access resources within
                         subfolders use forward slashes as separator.
        """
        if ('r', 'rb') not in mode:
            raise ValueError('Resources can only be opened for reading')
        return open(os.path.join(self.root_path, resource), mode)
dataset/ETHPy150Open baseblack/ReproWeb/3rdParty/python/flask/helpers.py/_PackageBoundObject.open_resource/swapped operands of `not in`