Fabric的connection.forward_local超出范围时失败

问题描述 投票:0回答:1

我正在尝试获取python脚本以启用从远程主机到本地计算机的端口转发以访问接口。

如果我手动执行ssh -L 54321:someotherhost:80 user@host(带有密码提示)可以正常工作,我可以按预期访问http://localhost:54321/someinterface上的界面。

现在我正在尝试使用织物:

from fabric import Connection

HOST = "somehost"
USER = "someuser"
PASSWORD = "somepassword"
LOCAL_PORT = "54321"
REMOTE_PORT = "80"
REMOTE_HOST = "someotherhost"

kwargs = {
    "password": PASSWORD
}
with Connection(HOST, user=USER, connect_kwargs=kwargs).forward_local(
        LOCAL_PORT, REMOTE_PORT, REMOTE_HOST, "localhost"
):
    pass # access interface e.g. via the requests package

但是,1.)转发似乎不起作用,原因未知,并且2.)在执行forward_local范围内的最后一行时,它停止并显示以下错误:

Traceback (most recent call last):
  File ".\path\to\script.py", line 67, in <module>
    main()
  File ".\path\to\script.py", line 35, in main
    pass
  File "C:\Users\ott\AppData\Local\Programs\Python\Python37\lib\contextlib.py", line 119, in __exit__
    next(self.gen)
  File "C:\Users\ott\AppData\Local\Programs\Python\Python37\lib\site-packages\fabric\connection.py", line 883, in forward_local
    raise ThreadException([wrapper])
invoke.exceptions.ThreadException: 
Saw 1 exceptions within threads (TypeError):

Thread args: {}

Traceback (most recent call last):

  File "C:\Users\ott\AppData\Local\Programs\Python\Python37\lib\site-packages\invoke\util.py", line 231, in run
    self._run()

  File "C:\Users\ott\AppData\Local\Programs\Python\Python37\lib\site-packages\fabric\tunnels.py", line 60, in _run
    sock.bind(self.local_address)

TypeError: an integer is required (got type str)

可能1.)和2.)相关,但是我现在专注于2.)。在forward_local生成的上下文管理器的范围内,我做什么都无所谓,在最后执行的语句上它停止。我认为这是由于当解释器离开作用域时,上下文管理器被python关闭时引起的。

python ssh fabric
1个回答
2
投票

根据documentation参数,例如:

  • LOCAL_PORT
  • REMOTE_PORT

必须是整数,而不是字符串。这就是为什么你得到:

TypeError:需要一个整数(got类型为str)

因此,更改变量:

LOCAL_PORT = "54321"
REMOTE_PORT = "80"

LOCAL_PORT = 54321
REMOTE_PORT = 80

应该解决问题。

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