[Python] Python Paramiko Filename needed on put command but

Environment: Windows DataCenter 2019.

I'm trying to send a file using Python/Paramiko to a non-standard sftp host. I believe it's a mainframe but I'm not sure. Destination filename needs to be, "//ABC.AB4.C3943.S46200_942145". That is the actual filename and it goes into the "root" folder because the remote host does the translation into whatever file system it needs. Behind the scenes, I suspect the dots are folders, but doesn't really matter.

What I'm getting from Paramiko appears to be the, "you must enter a destination filename" on the put statement error. I putting a filename. Just not what Paramiko expects?

Any ideas around this? Error log below... Code snippet below that...

Uploading \\xyz\tech\automation\temp\S46200_942145 to //ABC.AB4.C3943.S46200_942145
Traceback (most recent call last):
File "\\xyz\tech\Automation\scripts\dev\custom_jobs\Send_Files\generic_send_files.py", line 54, in <module>
r = sftp.put(source, destination)
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
File "c:\Automation\venv\Lib\site-packages\paramiko\sftp_client.py", line 759, in put
return self.putfo(fl, remotepath, file_size, callback, confirm)
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
File "c:\Automation\venv\Lib\site-packages\paramiko\sftp_client.py", line 714, in putfo
with self.file(remotepath, "wb") as fr:
^^^^^^^^^^^^^^^^^^^^^^^^^^^
File "c:\Automation\venv\Lib\site-packages\paramiko\sftp_client.py", line 372, in open
t, msg = self._request(CMD_OPEN, filename, imode, attrblock)
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
File "c:\Automation\venv\Lib\site-packages\paramiko\sftp_client.py", line 857, in _request
return self._read_response(num)
^^^^^^^^^^^^^^^^^^^^^^^^
File "c:\Automation\venv\Lib\site-packages\paramiko\sftp_client.py", line 909, in _read_response
self._convert_status(msg)
File "c:\Automation\venv\Lib\site-packages\paramiko\sftp_client.py", line 942, in _convert_status
raise IOError(text)
OSError: Failure


for files in os.listdir(source_path):
if fnmatch.fnmatch(files, source_pattern):
source = source_path + files
destination = remote_folder + files
print(f'Uploading {source} to {destination}')
r = sftp.put(source, destination)
print('Result:', r)


I don't know what to try.

I need the provided filename to be used/accepted by the put command.

Continue reading...