Issue
I have a private key file id_rsa
(starts with -----BEGIN RSA PRIVATE KEY-----). With the tool ssh-keygen
I am able to generate an SSH public key using the following command:
ssh-keygen -y -f ~/.ssh/id_rsa > ~/.ssh/id_rsa.pub
The generated file will have the following content:
ssh-rsa AAAAB3NzaC1yc2EAAAABIwAAAQEAklOUpkDHrfHY17SbrmTIpNLTGK9Tjom/BWDSU GPl+nafzlHDTYW7hdI4yZ5ew18JH4JW9j...
I am trying to achieve the same within my Python code. The code will be executed on AWS Lambda so I want to avoid invoking os.system()
to run shell commands since I don't have control over the underlying environment.
Given that I have a variable private_key
, how can I extract the ssh public key from it?
Solution
OpenSSH now has its own format for private keys (BEGIN OPENSSH PRIVATE KEY). Previously, the PKCS#1 or PKCS#8 format was used for private RSA keys. The posted private key has the PKCS#1 format (PEM encoded).
The Cryptography library supports a wide range of key formats, including PKCS#1 and the OpenSSH format. The following code allows importing a PKCS#1 formatted private key and exporting the public key in OpenSSH format:
from cryptography.hazmat.primitives import serialization
privatePkcs1Pem = b'''-----BEGIN RSA PRIVATE KEY-----
MIIEpAIBAAKCAQE....QkFn5HuC2aOZjktdA==
-----END RSA PRIVATE KEY-----'''
private_key = serialization.load_pem_private_key(privatePkcs1Pem, password=None)
public_key = private_key.public_key()
public_openssh = public_key.public_bytes(encoding=serialization.Encoding.OpenSSH, format=serialization.PublicFormat.OpenSSH )
print(public_openssh.decode('utf-8')) # ssh-rsa AAAAB3NzaC1...
Answered By - Topaco
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