Python 运行使用 Fabric 模块的 SSH 命令

Question

我正在尝试使用 Fabric 模块向两个 raspberry Pi 作为一个组发送运行命令。我正在尝试学习如何使用 Group 方法，但我认为我做错了什么，因为当我运行以下代码时：...

import fabric

b = fabric.connection.Connection("192.168.3.151", port=22, user="pi", \
    connect_kwargs={"password" : "Raspberry"})
c = fabric.connection.Connection("192.168.3.123", port=22, user="pi", \
    connect_kwargs={"password" : "Raspberry"})
pool = fabric.group.SerialGroup(b, c)
pool.run("touch /home/pi/Desktop/new_file65.txt")
pool.close()
print("hi")

我收到以下错误：

Traceback (most recent call last):
  File "/home/pi/.local/lib/python3.5/site-packages/invoke/config.py", line 98, in __getattr__
    return self._get(key)
  File "/home/pi/.local/lib/python3.5/site-packages/invoke/config.py", line 165, in _get
    value = self._config[key]
  File "/home/pi/.local/lib/python3.5/site-packages/invoke/config.py", line 154, in __getitem__
    return self._get(key)
  File "/home/pi/.local/lib/python3.5/site-packages/invoke/config.py", line 165, in _get
    value = self._config[key]
KeyError: 'rsplit'

During handling of the above exception, another exception occurred:

Traceback (most recent call last):
  File "test.py", line 7, in <module>
    pool = fabric.group.ThreadingGroup(b, c)
  File "/home/pi/.local/lib/python3.5/site-packages/fabric/group.py", line 70, in __init__
    self.extend(map(Connection, hosts))
  File "/home/pi/.local/lib/python3.5/site-packages/fabric/connection.py", line 246, in __init__
    shorthand = self.derive_shorthand(host)
  File "/home/pi/.local/lib/python3.5/site-packages/fabric/connection.py", line 416, in derive_shorthand
    user_hostport = host_string.rsplit("@", 1)
  File "/home/pi/.local/lib/python3.5/site-packages/invoke/config.py", line 110, in __getattr__
    raise AttributeError(err)
AttributeError: No attribute or config key found for 'rsplit'

Valid keys: ['connect_kwargs', 'forward_agent', 'gateway', 'load_ssh_configs', 'port', 'run', 'runners', 'ssh_config_path', 'sudo', 'tasks', 'timeouts', 'user']

Valid real attributes: ['cd', 'clear', 'client', 'close', 'config', 'connect_kwargs', 'connect_timeout', 'create_session', 'cwd', 'derive_shorthand', 'forward_agent', 'forward_local', 'forward_remote', 'from_data', 'gateway', 'get', 'host', 'is_connected', 'local', 'open', 'open_gateway', 'original_host', 'pop', 'popitem', 'port', 'prefix', 'put', 'resolve_connect_kwargs', 'run', 'setdefault', 'sftp', 'ssh_config', 'sudo', 'transport', 'update', 'user']

我想我犯了一个简单的错误，非常感谢任何有关分组连接的指导！谢谢！

Answer 1

这里有很多东西。要通过 fab 使用 fabfile.py 和运行命令，您必须确保使用 @task 修饰您的命令。这是一个例子：

--- 在 fabfile.py ---

from fabric.decorators import task
@task
def greet():
    print('Hello, Matt!')

当您想在 fabfile.py 内更改结构环境时（即，在 python 代码中的运行时间），您必须使用执行模式

from fabric.decorator import task
from fabric.operations import run
from fabric.context_managers import env
def touch_file():
    run("touch /home/pi/Desktop/new_file65.txt")

@task 
def manage_pis():
    env.hosts = [ ... ]
    execute(touch_file)

这应该足以作为使用 env 的起点。

Answer 2

SerialGroup 期望 string 而不是 Connection 所以你需要

import fabric

pool = fabric.group.SerialGroup("192.168.3.151", "192.168.3.123", user="pi", 
                                  port=22, connect_kwargs={"password": "Raspberry"})

pool.run("touch /home/pi/Desktop/new_file65.txt")
pool.close()
print("hi")

如果您需要不同的用户，您也可以使用login@host。但是你不能使用不同的密码。

pool = fabric.group.SerialGroup("pi@192.168.3.151", "pi@192.168.3.123", 
                                  port=22, connect_kwargs={"password": "Raspberry"})

文档：http://docs.fabfile.org/en/2.4/api/group.html#fabric.group.SerialGroup

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