Question

我正在编写custom state module，目的是在（部分）可配置内容的特定位置创建文件。

基本上我希望缩短以下SLS（为了这个问题而简化，实际上更复杂）......

/etc/foo/bar:
  file.managed:
    - contents: Hello World

......对此：

bar:
  my_module.foo:
    - message: World

由于此功能基本上是file.managed状态的更专业版本，因此在我的自定义状态模块中重用file.managed状态对我很有用。

有没有办法从我自己的状态模块调用file.managed状态模块？

我已经尝试过（不成功）：

导入salt.states.file并致电salt.states.file.managed：

import salt.states.file

def foo(name, message, **kwargs):
    return salt.states.file.managed(name='/etc/foo/%s' % name,
                                    contents='Hello %s' % message, 
                                    **kwargs)

这会导致错误消息：

  File "/usr/lib/python2.7/dist-packages/salt/state.py", line 1626, in call
    **cdata['kwargs'])
  File "/usr/lib/python2.7/dist-packages/salt/loader.py", line 1492, in wrapper
    return f(*args, **kwargs)
  File "/var/cache/salt/minion/extmods/states/my_module.py", line 14, in static_pod
    contents=yaml.dump(data, default_flow_style=False), **kwargs)
  File "/usr/lib/python2.7/dist-packages/salt/states/file.py", line 1499, in managed
    mode = __salt__['config.manage_mode'](mode)
NameError: global name '__salt__' is not defined

使用__salt__['file.managed']：

def foo(name, message, **kwargs):
    return __salt__['file.managed'](name='/etc/foo/%s' % name,
                                    contents='Hello %s' % message, 
                                    **kwargs)

这也导致（不同的）错误消息（这不足为奇，因为文档明确指出__salt__仅包含执行模块，而不是状态模块）：

  File "/usr/lib/python2.7/dist-packages/salt/state.py", line 1626, in call
    **cdata['kwargs'])
  File "/usr/lib/python2.7/dist-packages/salt/loader.py", line 1492, in wrapper
    return f(*args, **kwargs)
  File "/var/cache/salt/minion/extmods/states/my_module.py", line 13, in static_pod
    return __salt__['file.managed'](name='/etc/foo/%s' % name,
  File "/usr/lib/python2.7/dist-packages/salt/loader.py", line 900, in __getitem__
    func = super(LazyLoader, self).__getitem__(item)
  File "/usr/lib/python2.7/dist-packages/salt/utils/lazy.py", line 93, in __getitem__
    raise KeyError(key)
KeyError: 'file.managed'

Answer 1

我找到了使用state.single执行模块的解决方案：

def foo(name, messsage):
    result = __salt__['state.single'](fun='file.managed',
                                      name='/etc/foo/%s' % name,
                                      contents='Hello %s' % message,
                                      test=__opts__['test'])
    return result.items()[0][1]

state.single模块返回如下数据结构：

{
  'file_|-/etc/foo/bar_|-/etc/foo/bar_|-managed': {
    'comment': 'The file /etc/foo/bar is set to be changed',
    'name': '/etc/foo/bar',
    'start_time': '08:24:45.022518',
    'result': None,
    'duration': 18.019,
    '__run_num__': 0,
    'changes': {'diff': '...'}
  }
}

内部对象（file_|-/etc/foo/bar_|-/etc/foo/bar_|-managed下的所有内容）是file.managed模块自身返回的内容，因此您可以将此值重新用作自定义状态的返回值。

从状态模块内调用状态模块

1 个答案: