Question

如果使用puppet不存在，我想创建目录结构/var/www/apps/example/current/public。如果它已经存在，我不想清除目录的内容。我该怎么做呢？以下是我到目前为止：

file { "/var/www/apps/example/current/public":
  owner => 'deploy',
  group => 'users',
  ensure => "directory",
  purge => false,
  recurse => true
}

这给了我

 Cannot create /var/www/apps/example/current/public; parent directory /var/www/apps/example/current does not exist

Answer 1

recurse参数不允许您创建父目录。它用于递归地对目录内容和子目录强制执行owner，mode等属性值。

file { '/var/www':
    owner   => 'www-data',
    recurse => true,
}

事实上，Puppet目前无法自动创建所有父目录。您应该将所有相关目录添加为资源。

file { [ '/var/www/apps',
         '/var/www/apps/example',
         '/var/www/apps/example/current',
         '/var/www/apps/example/current/public', ]:
           ensure => directory,
           ...
}

现有内容将保持不受干扰。无需传递purge参数。

Answer 2

  exec { "Create ${mydir}":
    creates => $mydir,
    command => "mkdir -p ${mydir}",
    path => $::path
  } -> file { $mydir : }

最后一行是为了使其他资源（例如，在$mydir内创建的文件）可以依赖于File[$mydir]，就好像可以使用普通的旧file {}块创建它一样，which it really should。

Answer 3

木偶不会为你制作父目录，但你可以轻松创建一个类似文件的提供者。作为一个例子，我创建了一个自定义类型和提供程序，基本上运行＆＃39; mkdir -p＆＃39;在POSIX系统上：https://docs.puppetlabs.com/puppet/latest/reference/lang_namespaces.html

但是，有一个很好的理由，Puppet默认情况下不这样做。这是因为Puppet并不想对多个目录的权限负责，因为代理以root身份运行。如果您正在配置/ var / www或其他东西，这可能会很糟糕。

recuse file参数实际上是用来管理目录树参数：https://docs.puppetlabs.com/references/latest/type.html#file-attribute-recurse

您可以使用source =＆gt;创建目录树并提供服务。＆＃39;木偶：///＆＃39;例如，将recurse设置为true，并且它将使用在所服务的目录树上设置的所有文件模式。

Answer 4

我试图找到一个好的解决方案，但失败了。所以我自己想出了办法。希望它对其他人有用。

以下函数将生成父目录列表，然后我们可以使用该列表来构建父文件夹。第一个参数是用作搜索父目录的起点的路径;第二个参数是可选的，它用作终止点（不包括）来停止查找：

module Puppet::Parser::Functions
  newfunction(:parentdirs, :type => :rvalue, :doc => <<-EOS
    Build a list of all its parent directories.
    EOS
  ) do |arguments|

    raise(Puppet::ParseError, "parentdirs(): Wrong number of arguments " +
      "given (#{arguments.size} for 1)") if arguments.size < 1

    $dir_until = arguments.size > 1 ? arguments[1] : nil
    $cur = File.dirname(arguments[0])
    $result = []
    begin
        $result.unshift($cur)
        $last = $cur
        $cur = File.dirname($cur)
    end while $cur != $last and !$cur.end_with?('/') and $cur != $dir_until

    return $result
  end
end

以下是如何使用它的示例：

$my_folder = '/var/www/apps/example/current/public'
$parent_dirs = parentdirs($my_folder, '/var/www/apps')
file { $parent_dirs:
  ensure => 'directory',
  owner => 'deploy',
  group => 'users'
}
file { $my_folder:
  ensure => 'directory',
  owner => 'deploy',
  group => 'anonymous'
}

上述代码将确保文件夹＆＃39; / var / www / apps / example＆＃39;和＆＃39; / var / www / apps / example / current＆＃39;在创建＆var; / var / www / apps / example / current / public＆＃39;之前创建而＆＃39; / var / www / apps / example＆＃39;以上仍然没有被触及。

我只在Windows中测试过它。但它应该在Linux环境中工作。

这不太理想。但它比手动逐一列出所有父母更好。

Answer 5

如果您使用＆＃34; define＆＃34;，您可以使用以下内容：

mymodule::recursive_dir { "My Directory" :
   drive => "C:",
   path  => "/path/to/folder",
}

我定义＆＃34;定义＆＃34;在mymodule.rb中：

define mymodule::recursive_dir ($drive, $path) {
  $folders = split($path, "/")
  $folders.each |$index, $folder| {
    $calculated_folder = inline_template("<%= @folders[0, @index + 1].join('/') %>")
    $full_path = "${drive}${calculated_folder}"
    if (! defined(File[$full_path]) and $full_path != $drive) {
      file { $full_path :
        ensure => directory,
      }
    }
  }
}

这会拆分路径并创建每个目录，因为它会将路径重新组合在一起，确保不要尝试自己创建驱动器。

Answer 6

Thank-you to AnthonyY.

I didn't know where to put the code for his function, and as a result of my research ended up rewriting the whole thing using the newer syntax, but keeping his logic.

It took me quite a while to work it all out, so I figured it would be good to post it back here. ...I would have just added it as a comment to his answer, but apparently I need 50 points to do that, not zero.

The code should be saved in it's own file inside your Puppet environment directory as follows:

lib/puppet/functions/parentdirs.rb

...so the full path would be something like this (on Ubuntu server 18.04, using the Puppet packages, not the repo ones):

/etc/puppetlabs/code/environments/testing/lib/puppet/functions/parentdirs.rb

...there seem to be other places you can put it, but this is what I did.

Notice that the file is .rb, not .pp (because it's Ruby code, not Puppet).

I got most of my information from https://puppet.com/docs/puppet/5.5/functions_ruby_overview.html and the sub-pages.

The usage is the same as the original function, and somewhat re-explained in the comments

# Returns an array of the parent directories to the given file or directory. This can then be passed to File to create the directory tree require for a dynamic path value.
# Parameter 2 is an optional, higher level of the same path. These higher level directories will not be in the array.
# Example 1: parameter 1 is '/var/www/mysite'; parameter 2 is not given; returns array ['/var', '/var/www']
# Example 2: parameter 1 is '/var/www/mysite'; parameter 2 is '/var'; returns array ['/var/www']

Puppet::Functions.create_function(:parentdirs) do

  dispatch :parents do
    required_param 'String', :target_dir
    optional_param 'String', :dir_until
    return_type 'Array'
  end


  def parents(target_dir, dir_until = '')

    cur = File.dirname(target_dir)
    result = []

    begin
        result.unshift(cur)
        last = cur
        cur = File.dirname(cur)
    end while cur != last and !cur.end_with?('/') and cur != dir_until

    return result

  end

end

Answer 7

这是做mkdir -p $(dirname $file_path)的纯木偶解决方案

$file_path = '/tmp/foo/bar/bob.conf' # assumes file_path is Stdlib::Unixpath
# strip leading '/' then split and loop
$dirs = $file_path[1,-1].dirname.split('/').reduce([]) |$memo, $subdir| {
    $_dir =  $memo.empty ? { 
        true    => "/${subdir}",
        default => "${$memo[-1]}/${subdir}",
    }     
    concat($memo, $_dir)
}
file {$dirs:
    ensure => directory,
}

递归地在puppet中制作目录树而不进行清除

7 个答案: