Puppet:文件资源仅当文件存在时 [英] Puppet : file resource only if file exists

问题描述

我想做的很简单:

1.

将 /source/file 复制到 /target/file.我使用以下方法实现了这一点:

Copy /source/file to /target/file. I achieve this using the following:

file { 'my_file_copy':
  ensure   => file,
  source   => 'file:/source/file',
  path     => "/target/file",
}

2.

但是，如果文件 /source/file 不存在，我不希望它执行此任务.

However, if file /source/file does not exist, I do NOT want it to perform this task.

我真的很纠结这个逻辑.我尝试了下面的解决方案，但它在人偶运行期间抛出异常.

I am really struggling with this logic. I attempted the solution below but it throws exceptions during puppet run.

puppet:如果一个文件存在则复制另一个文件结束

有没有更好的方法来完成这项任务?理想情况下，我只想使用文件"而避免使用exec".但在这一点上，我会满足于一个解决方案！

Is there a better way of achieving this task ? Ideally, I would like to only use "file" and avoid using "exec". But at this point I would settle for a solution !

推荐答案

因为 Puppet 是一种声明式语言，其中只声明结束状态，所以命令式逻辑(如您所描述的 - 如果 A，执行 X - 通常是难以表达.

Because Puppet is a declarative language where only the end-state is declared, imperative logic such as what you've described - if A, do X - is often hard to express.

就我个人而言，当且仅当文件 A 存在时，我会尽量避免复制文件 B 的要求.通常有更好的方法.

Personally, I would try to simply avoid this requirement of having file B copied if and only if file A exists. Often there's a better way.

但是，如果需要保留要求，那么在这里使用 Exec 对我来说是一个不错的选择.

If the requirement needs to stay, however, then use of Exec here sounds like a pretty good option to me.

exec { 'my_file_copy':
  command => 'cp /source/file /target/file',
  onlyif  => 'test -e /source/file',
  creates => '/target/file',
  path    => '/bin',
}

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