...
I tried an experiment with a remote server that I control and pysftp works perfectly there. A difference that I know of is that
this server is using ubuntu 18.04 and we don't use passwords, but a private_key. Also this server is using the openssh internal sftp.
I believe the failing se
On 13/06/2019 18:23, MRAB wrote:
.
What does:
sftp.normalize('.')
return?
It returns '/'.
sftp.chdir('') and that also fails in paramiko as it seems to use
CMD_REALPATH to do that.
File "tsftp.py", line 7, in main
print(sftp.chdir(''))
File "/home/rptlab/tmp/tpy3
On 2019-06-13 14:57, Robin Becker wrote:
On 13/06/2019 05:56, dieter wrote:
Robin Becker writes:
On 12/06/2019 05:59, dieter wrote:
Robin Becker writes:
I am trying to convert older code that uses ftplib as the endpoint has switched
to sftp only.
...
Well with real sftp I can cd to that p
On 13/06/2019 05:56, dieter wrote:
Robin Becker writes:
On 12/06/2019 05:59, dieter wrote:
Robin Becker writes:
I am trying to convert older code that uses ftplib as the endpoint has switched
to sftp only.
...
Well with real sftp I can cd to that path so if it is a symlink it goes
somewhe
Robin Becker writes:
> On 12/06/2019 05:59, dieter wrote:
>> Robin Becker writes:
>>> I am trying to convert older code that uses ftplib as the endpoint has
>>> switched to sftp only.
> ...
> Well with real sftp I can cd to that path so if it is a symlink it goes
> somewhere.
>
> With pysftp I
On 12/06/2019 05:59, dieter wrote:
Robin Becker 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.rem
Robin Becker 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'
