(Chilkat2-Python) SFTP Upload from Local Filesystem

Demonstrates how to upload a file to an SSH server. There are no limitations on file size, and the file is streamed directly from the local filesystem.

Chilkat2 Python Downloads install with pip pip3 install chilkat2 or download... Python Module for Windows, Linux, Alpine Linux, MacOS, Solaris

import sys
import chilkat2

# This example assumes the Chilkat API to have been previously unlocked.
# See Global Unlock Sample for sample code.

sftp = chilkat2.SFtp()

# Set some timeouts, in milliseconds:
sftp.ConnectTimeoutMs = 5000
sftp.IdleTimeoutMs = 10000

# Connect to the SSH server.  
# The standard SSH port = 22
# The hostname may be a hostname or IP address.
hostname = "sftp.example.com"
port = 22
success = sftp.Connect(hostname,port)
if (success != True):
    print(sftp.LastErrorText)
    sys.exit()

# Authenticate with the SSH server.  Chilkat SFTP supports
# both password-based authenication as well as public-key
# authentication.  This example uses password authenication.
success = sftp.AuthenticatePw("myLogin","myPassword")
if (success != True):
    print(sftp.LastErrorText)
    sys.exit()

# After authenticating, the SFTP subsystem must be initialized:
success = sftp.InitializeSftp()
if (success != True):
    print(sftp.LastErrorText)
    sys.exit()

# Open a file for writing on the SSH server.
# If the file already exists, it is overwritten.
# (Specify "createNew" instead of "createTruncate" to
# prevent overwriting existing files.)
handle = sftp.OpenFile("hamlet.xml","writeOnly","createTruncate")
if (sftp.LastMethodSuccess != True):
    print(sftp.LastErrorText)
    sys.exit()

# Upload from the local file to the SSH server.
success = sftp.UploadFile(handle,"c:/temp/hamlet.xml")
if (success != True):
    print(sftp.LastErrorText)
    sys.exit()

# -----------------------------------------------------------------------------
# This is important.  You must close the handle on the server.
# Otherwise open handles will accumulate on the server until eventually a limit
# is reached and the server will fail on a call to OpenFile.
# -----------------------------------------------------------------------------

# Close the file.
success = sftp.CloseHandle(handle)
if (success != True):
    print(sftp.LastErrorText)
    sys.exit()

print("Success.")