pysftp / paramiko problem
dieter
dieter at handshake.de
Wed Jun 12 00:59:17 EDT 2019
Robin Becker <robin at reportlab.com> writes:
> I am trying to convert older code that uses ftplib as the endpoint has switched to sftp only.
>
> I am using the pysftp wrapper around paramiko.
>
> The following script fails
>
> def main():
> import pysftp
> with pysftp.Connection('ftp.remote.com', username='me', password='xxxxxx') as sftp:
> print('top level')
> print(sftp.listdir())
> print(sftp.normalize(u'XXXX'))
>From the "sftp" documentation:
| normalize(self, remotepath)
| Return the expanded path, w.r.t the server, of a given path. This
| can be used to resolve symlinks or determine what the server believes
| to be the :attr:`.pwd`, by passing '.' as remotepath.
This suggests that your observation could be explained
by "u'XXXX'" being a broken symlink.
More information about the Python-list
mailing list