Upload file via SFTP with Python

Question

I wrote a simple code to upload a file to a SFTP server in Python. I am using Python 2.7.

import pysftp

srv = pysftp.Connection(host="www.destination.com", username="root",
password="password",log="./temp/pysftp.log")

srv.cd('public') #chdir to public
srv.put('C:\Users\XXX\Dropbox\test.txt') #upload file to nodejs/

# Closes the connection
srv.close()

The file did not appear on the server. However, no error message appeared. What is wrong with the code?

I have enabled logging. I discovered that the file is uploaded to the root folder and not under public folder. Seems like srv.cd('public') did not work.

Brew

Answer 1

I found the answer to my own question.

import pysftp

srv = pysftp.Connection(host="www.destination.com", username="root",
password="password",log="./temp/pysftp.log")

with srv.cd('public'): #chdir to public
    srv.put('C:\Users\XXX\Dropbox\test.txt') #upload file to nodejs/

# Closes the connection
srv.close()

Put the srv.put inside with srv.cd

Brew

Pysftp is dead. Better use Paramiko. See my answer. – Nonrestrictive

Answer 2

Do not use pysftp it's dead. Use Paramiko directly. See also pysftp vs. Paramiko.

---

The code with Paramiko will be pretty much the same, except for the initialization part.

import paramiko
 
with paramiko.SSHClient() as ssh:
    ssh.load_system_host_keys()
    ssh.connect(host, username=username, password=password)
 
    sftp = ssh.open_sftp()

    sftp.chdir('public')
    sftp.put('C:\Users\XXX\Dropbox\test.txt', 'test.txt')

---

To answer the literal OP's question: the key point here is that pysftp Connection.cd works as a context manager (so its effect is discarded without with statement), while Paramiko SFTPClient.chdir does not.

Nonrestrictive

Answer 3

import pysftp

with pysftp.Connection(host="www.destination.com", username="root",
password="password",log="./temp/pysftp.log") as sftp:

  sftp.cwd('/root/public')  # The full path
  sftp.put('C:\Users\XXX\Dropbox\test.txt')  # Upload the file

No sftp.close() is needed, because the connection is closed automatically at the end of the with-block

I did a minor change with cd to cwd

Syntax -

# sftp.put('/my/local/filename')  # upload file to public/ on remote
# sftp.get('remote_file')         # get a remote file

Antigen

The "minor change" of cd to cwd it the actual fix. The cd works as a context manager, while cwd does not. – Nonrestrictive