尽管这不使用 paramiko,但我相信它是一个非常干净的实现解决方案(类似于 @dario's answer,但无需在 python 中管理线程)。
openssh 客户端中有一个鲜为人知的功能,它允许我们通过 unix 套接字控制 ssh 进程,引用 man ssh:
-M Places the ssh client into “master” mode for connection sharing. Multiple -M options places ssh
into “master” mode with confirmation required before slave connections are accepted. Refer to the
description of ControlMaster in ssh_config(5) for details.
-S ctl_path
Specifies the location of a control socket for connection sharing, or the string “none” to disable
connection sharing. Refer to the description of ControlPath and ControlMaster in ssh_config(5)
for details.
因此您可以启动ssh(使用-Nf)的后台进程,然后使用另一个ssh 调用来检查(或终止)它。
我在需要建立反向隧道的项目中使用它
from subprocess import call, STDOUT
import os
DEVNULL = open(os.devnull, 'wb')
CONFIG = dict(
SSH_SERVER='ssh.server.com',
SSH_PORT=2222,
SSH_USER='myuser',
SSH_KEY='/path/to/user.key',
REMOTE_PORT=62222,
UNIX_SOCKET='/tmp/ssh_tunnel.sock',
KNOWN_HOSTS='/path/to/specific_known_host_to_conflicts',
)
def start():
return call(
[
'ssh', CONFIG['SSH_SERVER'],
'-Nfi', CONFIG['SSH_KEY'],
'-MS', CONFIG['UNIX_SOCKET'],
'-o', 'UserKnownHostsFile=%s' % CONFIG['KNOWN_HOSTS'],
'-o', 'ExitOnForwardFailure=yes',
'-p', str(CONFIG['SSH_PORT']),
'-l', CONFIG['SSH_USER'],
'-R', '%d:localhost:22' % CONFIG['REMOTE_PORT']
],
stdout=DEVNULL,
stderr=STDOUT
) == 0
def stop():
return __control_ssh('exit') == 0
def status():
return __control_ssh('check') == 0
def __control_ssh(command):
return call(
['ssh', '-S', CONFIG['UNIX_SOCKET'], '-O', command, 'x'],
stdout=DEVNULL,
stderr=STDOUT
)
-o ExitOnForwardFailure=yes确保隧道无法建立时ssh命令会失败,否则不会退出。