SFTP in Python? (platform independent)

Question:

I’m working on a simple tool that transfers files to a hard-coded location with the password also hard-coded. I’m a python novice, but thanks to ftplib, it was easy:

import ftplib

info= ('someuser', 'password')    #hard-coded

def putfile(file, site, dir, user=(), verbose=True):
    """
    upload a file by ftp to a site/directory
    login hard-coded, binary transfer
    """
    if verbose: print 'Uploading', file
    local = open(file, 'rb')    
    remote = ftplib.FTP(site)   
    remote.login(*user)         
    remote.cwd(dir)
    remote.storbinary('STOR ' + file, local, 1024)
    remote.quit()
    local.close()
    if verbose: print 'Upload done.'

if __name__ == '__main__':
    site = 'somewhere.com'            #hard-coded
    dir = './uploads/'                #hard-coded
    import sys, getpass
    putfile(sys.argv[1], site, dir, user=info)

The problem is that I can’t find any library that supports sFTP. What’s the normal way to do something like this securely?

Edit: Thanks to the answers here, I’ve gotten it working with Paramiko and this was the syntax.

import paramiko

host = "THEHOST.com"                    #hard-coded
port = 22
transport = paramiko.Transport((host, port))

password = "THEPASSWORD"                #hard-coded
username = "THEUSERNAME"                #hard-coded
transport.connect(username = username, password = password)

sftp = paramiko.SFTPClient.from_transport(transport)

import sys
path = './THETARGETDIRECTORY/' + sys.argv[1]    #hard-coded
localpath = sys.argv[1]
sftp.put(localpath, path)

sftp.close()
transport.close()
print 'Upload done.'

Thanks again!

Asked By: Mark Wilbur

||

Answers:

Twisted can help you with what you are doing, check out their documentation, there are plenty of examples. Also it is a mature product with a big developer/user community behind it.

Answered By: Sergey Golovchenko

Paramiko supports SFTP. I’ve used it, and I’ve used Twisted. Both have their place, but you might find it easier to start with Paramiko.

Answered By: Brian Clapper

If you want easy and simple, you might also want to look at Fabric. It’s an automated deployment tool like Ruby’s Capistrano, but simpler and of course for Python. It’s build on top of Paramiko.

You might not want to do ‘automated deployment’ but Fabric would suit your use case perfectly none the less. To show you how simple Fabric is: the fab file and command for your script would look like this (not tested, but 99% sure it will work):

fab_putfile.py:

from fabric.api import *

env.hosts = ['THEHOST.com']
env.user = 'THEUSER'
env.password = 'THEPASSWORD'

def put_file(file):
    put(file, './THETARGETDIRECTORY/') # it's copied into the target directory

Then run the file with the fab command:

fab -f fab_putfile.py put_file_file=./path/to/my/file

And you’re done! 🙂

Answered By: hopla

You should check out pysftp https://pypi.python.org/pypi/pysftp it depends on paramiko, but wraps most common use cases to just a few lines of code.

import pysftp
import sys

path = './THETARGETDIRECTORY/' + sys.argv[1]    #hard-coded
localpath = sys.argv[1]

host = "THEHOST.com"                    #hard-coded
password = "THEPASSWORD"                #hard-coded
username = "THEUSERNAME"                #hard-coded

with pysftp.Connection(host, username=username, password=password) as sftp:
    sftp.put(localpath, path)

print 'Upload done.'
Answered By: Dundee MT

Here is a sample using pysftp and a private key.

import pysftp

def upload_file(file_path):

    private_key = "~/.ssh/your-key.pem"  # can use password keyword in Connection instead
    srv = pysftp.Connection(host="your-host", username="user-name", private_key=private_key)
    srv.chdir('/var/web/public_files/media/uploads')  # change directory on remote server
    srv.put(file_path)  # To download a file, replace put with get
    srv.close()  # Close connection

pysftp is an easy to use sftp module that utilizes paramiko and pycrypto. It provides a simple interface to sftp.. Other things that you can do with pysftp which are quite useful:

data = srv.listdir()  # Get the directory and file listing in a list
srv.get(file_path)  # Download a file from remote server
srv.execute('pwd') # Execute a command on the server

More commands and about PySFTP here.

Answered By: radtek

There are a bunch of answers that mention pysftp, so in the event that you want a context manager wrapper around pysftp, here is a solution that is even less code that ends up looking like the following when used

path = "sftp://user:p@[email protected]/path/to/file.txt"

# Read a file
with open_sftp(path) as f:
    s = f.read() 
print s

# Write to a file
with open_sftp(path, mode='w') as f:
    f.write("Some content.") 

The (fuller) example: http://www.prschmid.com/2016/09/simple-opensftp-context-manager-for.html

This context manager happens to have auto-retry logic baked in in the event you can’t connect the first time around (which surprisingly happens more often than you’d expect in a production environment…)

The context manager gist for open_sftp: https://gist.github.com/prschmid/80a19c22012e42d4d6e791c1e4eb8515

Answered By: prschmid

Paramiko is so slow. Use subprocess and shell, here is an example:

remote_file_name = "filename"
remotedir = "/remote/dir"
localpath = "/local/file/dir"
    ftp_cmd_p = """
    #!/bin/sh
    lftp -u username,password sftp://ip:port <<EOF
    cd {remotedir}
    lcd {localpath}
    get {filename}
    EOF
    """
subprocess.call(ftp_cmd_p.format(remotedir=remotedir,
                                 localpath=localpath,
                                 filename=remote_file_name 
                                 ), 
                shell=True, stdout=sys.stdout, stderr=sys.stderr)
Answered By: 杨李思

With RSA Key then refer here

Snippet:

import pysftp
import paramiko
from base64 import decodebytes

keydata = b"""AAAAB3NzaC1yc2EAAAADAQABAAABAQDl""" 
key = paramiko.RSAKey(data=decodebytes(keydata)) 
cnopts = pysftp.CnOpts()
cnopts.hostkeys.add(host, 'ssh-rsa', key)


with pysftp.Connection(host=host, username=username, password=password, cnopts=cnopts) as sftp:   
  with sftp.cd(directory):
    sftp.put(file_to_sent_to_ftp)
Answered By: Abhijeet

PyFilesystem with its sshfs is one option. It uses Paramiko under the hood and provides a nicer paltform independent interface on top.

import fs

sf = fs.open_fs("sftp://[user[:password]@]host[:port]/[directory]")
sf.makedir('my_dir')

or

from fs.sshfs import SSHFS
sf = SSHFS(...
Answered By: fmalina

fsspec is a great option for this, it offers a filesystem like implementation of sftp.

from fsspec.implementations.sftp import SFTPFileSystem
fs = SFTPFileSystem(host=host, username=username, password=password)

# list a directory
fs.ls("/")

# open a file
with fs.open(file_name) as file:
    content = file.read()

Also worth noting that fsspec uses paramiko in the implementation.

Answered By: Brian Larsen

Here’s a generic function that will download any given sftp url to a specified path

from urllib.parse import urlparse
import paramiko

url = 'sftp://username:password@hostname/filepath.txt'

def sftp_download(url, dest):
    url = urlparse(url)
    with paramiko.Transport((url.hostname, 22)) as transport:
        transport.connect(None,url.username,url.password)
        with paramiko.SFTPClient.from_transport(transport) as sftp:
            sftp.get(url.path, dest)

Call it with

sftp_download(url, "/tmp/filepath.txt")
Answered By: Jonathan
Categories: questions Tags: ,
Answers are sorted by their score. The answer accepted by the question owner as the best is marked with
at the top-right corner.