Python PYSFTP - pass private-key as string/text instead of passing file path

Question

I would like to pass my actual private-key value as argument instead of providing the file path.

I have used below code as of now:

import pysftp
import os

cnopts = pysftp.CnOpts()
if str(host_keys).lower() =='none':
    cnopts.hostkeys = None
else:
    cnopts.hostkeys.load(hostkeys)
filename = os.path.basename(localpath)
print(filename)
remotepath = os.path.join(remotefolder, filename)
print(remotepath)
with pysftp.Connection(host=hostname, port=int(port), username=username, password=password, cnopts=cnopts,private_key=private_key_filepath) as sftp:
    sftp.put(localpath, remotepath=remotepath)

Please suggest some way to pass it as text.

Example:

private_key='abcdmyprivatekeytext'

In actual scenario I will be placing my private-key text in secure vault.

Answer 1

The pysftp can accept RSAKey in the private_key argument of Connection constructor:

# Set your private key as a string
PRIVATE_KEY = """-----BEGIN RSA PRIVATE KEY-----
...
-----END RSA PRIVATE KEY-----
"""

# Use 'io.StringIO' to read your string as a file-like object.
privkey = io.StringIO(PRIVATE_KEY)

# Use paramiko to create your RSAKey
ki = paramiko.RSAKey.from_private_key(privkey)

# Connect using your key with pysftp
conn = pysftp.Connection(host=HOST, username=USER, private_key=ki)

(originally posted by @Hana, but the answer is deleted now)

This use of private_key argument is not backed by documentation as of pysftp 0.2.9, but it works. Other key types (DSSKey, ECDSAKey, Ed25519Key) are not accepted.

---

If you need to use other keys types, use Paramiko directly:
SSH/SCP through Paramiko with key in string

The pysftp is just an abandoned wrapper around Paramiko.
Prefer using Paramiko directly: pysftp vs. Paramiko