Compute syncthing device ID from its certificate

syncthing_id.py

#!/usr/bin/env python
import base64
import hashlib
import subprocess
import sys

B32ALPHABET = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ234567'

def luhn_checksum(data, alphabet=B32ALPHABET):
    n = len(alphabet)
    number = tuple(alphabet.index(i) for i in reversed(data))
    result = (sum(number[::2]) +
              sum(sum(divmod(i * 2, n)) for i in number[1::2])) % n
    return alphabet[-result]

def main(incert):
    der_data = subprocess.check_output(['openssl', 'x509', '-outform', 'DER'], stdin=incert)
    data_hash = hashlib.sha256(der_data)
    b32_hash = base64.b32encode(data_hash.digest()).decode('ascii')

    result = b32_hash.upper().rstrip('=')
    blocks = [result[pos:pos+13] for pos in range(0, len(result), 13)]
    result = ''.join(block + luhn_checksum(block) for block in blocks)

    blocks = [result[pos:pos+7] for pos in range(0, len(result), 7)]
    print('-'.join(blocks))

if __name__ == '__main__':
    import argparse
    parser = argparse.ArgumentParser(description='Generate syncthing ID from certificate')
    parser.add_argument('incert', type=argparse.FileType('rb'), help='Certificate path')
    args = parser.parse_args()
    main(**vars(args))

Usage example: python syncthing_id.py ~/.config/syncthing/cert.pem
You can pass - as a filename to make it read the certificate from its standard input.
Writes the corresponding syncthing device ID on its standard ouput.

No dependencies. Works with both python2 and python3.
Tested on Debian and Ubuntu. No idea if it would work on Windows.

It wouldn't most likely (due to openssl), and it does depend on openssl being installed, as well as requires the cert to be in DER format

Fair enough, as openssl comes in most linux default installs, I forgot to count it as a dependency.
Certificate can be in any format, the point of using the openssl command is to convert it from whatever format into DER.

Using ssl.PEM_cert_to_DER_cert removes the openssl dependency and makes this work on Windows

@spectras Thanks for sharing this, it was a big help to me.

Regards,
iain