我正在尝试理解获取OpenSSH公钥的步骤,如下所示:
rsa ssh- AAAAB3NzaC1yc2EAAAABIwAAAQEAqmEmDTNBC6O8HGCdu0MZ9zLCivDsYSttrrmlq87/YsEBpvwUTiF3UEQuFLaq5Gm+dtgxJewg/UwsZrDFxzpQhCHB6VmqrbKN2hEIkk/HJvCnAmR1ehXv8n2BWw3Jlw7Z+VgWwXAH50f2HWYqTaE4qP4Dxc4RlElxgNmlDPGXw/dYBvChYBG/RvIiTz1L+pYzPD4JR54IMmTOwjcGIJl7nk1VjKvl3D8Wgp6qejv4MfZ7Htdc99SUKcKWAeHYsjPXosSk3GlwKiS/sZi51Yca394GE7T4hZu6HTaXeZoD8+IZ7AijYn89H7EPjuu0iCAa/cjVzBsFHGszQYG+U5KfIw==
然后将其转换为标准指纹,如下所示:
2048 49:d3:cb:f6:00:d2:93:43:a6:27:07:ca:12:fd:5d:98 id_rsa.pub (RSA)
我试图深入OpenSSH源代码来理解这一点,但这超出了我的能力范围。我的第一个猜测是对关键文本执行一个简单的MD5,但结果与上面的输出不匹配。
发布于 2011-07-14 01:19:29
它是base64编码密钥的MD5和:
import base64
import hashlib
def lineToFingerprint(line):
key = base64.b64decode(line.strip().split()[1].encode('ascii'))
fp_plain = hashlib.md5(key).hexdigest()
return ':'.join(a+b for a,b in zip(fp_plain[::2], fp_plain[1::2]))
发布于 2014-12-18 22:15:44
https://github.com/ojarva/sshpubkeys
pip install sshpubkeys
用法:
import sshpubkeys
key = sshpubkeys.SSHKey("ssh-rsa AAAAB3NzaC1yc2EAAAABIwAAAQEAqmEmDTNBC6O8H" +
"GCdu0MZ9zLCivDsYSttrrmlq87/YsEBpvwUTiF3UEQuFLaq5Gm+dtgxJewg/UwsZrDFxz" +
"pQhCHB6VmqrbKN2hEIkk/HJvCnAmR1ehXv8n2BWw3Jlw7Z+VgWwXAH50f2HWYqTaE4qP4" +
"Dxc4RlElxgNmlDPGXw/dYBvChYBG/RvIiTz1L+pYzPD4JR54IMmTOwjcGIJl7nk1VjKvl" +
"3D8Wgp6qejv4MfZ7Htdc99SUKcKWAeHYsjPXosSk3GlwKiS/sZi51Yca394GE7T4hZu6H" +
"TaXeZoD8+IZ7AijYn89H7EPjuu0iCAa/cjVzBsFHGszQYG+U5KfIw== user@host")
print(key.bits) # 2048
print(key.hash()) # '49:d3:cb:f6:00:d2:93:43:a6:27:07:ca:12:fd:5d:98'
https://stackoverflow.com/questions/6682815
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