I found two issues with the put_d() function.

1. It looks like it doesn't accept absolute Windows paths. I get the error:

ValueError: 'C:\\Users\\johnm\\PycharmProjects\\project\\.other\\folder\\file' does not start with '\\'

2. When I call put_d("local_folder", "folder"), I get the following error:

File "C:\Users\johnm\PycharmProjects\project\venv\lib\site-packages\sftpretty\__init__.py", line 563, in put confirm=confirm, preserve_mtime=preserve_mtime) File "C:\Users\johnm\PycharmProjects\project\venv\lib\site-packages\sftpretty\__init__.py", line 554, in _put confirm=confirm) File "C:\Users\johnm\PycharmProjects\project\venv\lib\site-packages\paramiko\sftp_client.py", line 759, in put return self.putfo(fl, remotepath, file_size, callback, confirm) File "C:\Users\johnm\PycharmProjects\project\venv\lib\site-packages\paramiko\sftp_client.py", line 714, in putfo with self.file(remotepath, "wb") as fr: File "C:\Users\johnm\PycharmProjects\project\venv\lib\site-packages\paramiko\sftp_client.py", line 372, in open t, msg = self._request(CMD_OPEN, filename, imode, attrblock) File "C:\Users\johnm\PycharmProjects\project\venv\lib\site-packages\paramiko\sftp_client.py", line 822, in _request return self._read_response(num) File "C:\Users\johnm\PycharmProjects\project\venv\lib\site-packages\paramiko\sftp_client.py", line 874, in _read_response self._convert_status(msg) File "C:\Users\johnm\PycharmProjects\project\venv\lib\site-packages\paramiko\sftp_client.py", line 903, in _convert_status raise IOError(errno.ENOENT, text) FileNotFoundError: [Errno 2] Path not found.

As I understand, the path it refers to is the remote path. When I debug it, in put_d(), the paths it creates are the following: As you can see, the remotedir "folder/local_folder/file" is probably the issue. In my remote location, it creates the folder "folder". This of course makes sense as it successfully calls self.mkdir_p(remotedir) in line 599. Do you think this has to do with the location of the self.mkdir_p(remotedir) call? Should there be code that also creates the subfolders of the "local_folder" in the remote location(here the "folder/local_folder" subfolder)?

Thank you for your time.

Certain SFTP server do not allow to open a channel for every command. The original pysftp used a single channel for all actions. I've added a flag to allow us to work in a similar way, while by default keeping the new behavior.

Those servers will simply close the connection on the second command issued and it makes sftpretty not very useful with these old(er) SFTP servers.

Hi,

In sftpretty/init.py:716 there is a typo in the tries argument which causes the call to putfo() to raise an exception

retry() got an unexpected keyword argument 'tires'

Sometimes when assigning to _channel, an exception can occur in SFTPClient.from_transport(). Because _channel doesn't exist if that happens, it causes a second exception when trying to call close() in the finally block. Assign to the variable outside the try block initially so that this exception can't happen in the future. Also make sure that _channel is not None before we try to close it to account for it being sometimes None.

Calling basicConfig on the top of the file for a library that is embedded into other projects can cause issue. I suggest you drop it completely specifically because there are no change to it other than setting the default level to INFO.

https://github.com/byteskeptical/sftpretty/blob/ca8e2414ac156276059fcd091cc91e489d856e7d/sftpretty/init.py#L22

In our case it added ":" to the start of every log line.