pysftp / paramiko problem
On 12/06/2019 05:59, dieter wrote:
> 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')
> 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.
Well with real sftp I can cd to that path so if it is a symlink it goes somewhere.
With pysftp I am unable to chdir or cd into it. With a bit of difficulty I can use subprocess + sshpass + sftp to do the required