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我在木偶中有以下代码

if $::operatingsystemmajrelease < 7 {

     file { '/etc/sudoers' :
      ensure  => present,
      owner   => 'root',
      group   => 'root',
      mode    => '0440',
      content => template('sudo/sudoers.erb'),
    }
  }

我为此编写了 serverspec

require 'spec_helper'

if os[:release] == '6'

  describe file('/etc/sudoers') do
    it { should exist }
    it { should be_owned_by 'root' }
    it { should be_grouped_into 'root' }
    it { should be_mode 440 }
    its(:content) { should match /Defaults    requiretty/}
  end
end

1）为什么我收到以下消息

No examples found.

Finished in 0.00038 seconds (files took 0.44176 seconds to load)
0 examples, 0 failures

我的操作系统版本是 6，所以它应该编译我的 serverspec 代码，但为什么它无法执行所有示例。

如果我将服务器规范更改为if os[:release] < '7'

2)如果我给} 4 个数字八位字节匹配it { should be_mode 440 }，也不支持。it { should be_mode 0440

低于错误

File "/etc/sudoers" should be mode 288
     On host `localhost'
     Failure/Error: it { should be_mode 0440 }
       expected `File "/etc/sudoers".mode?(288)` to return true, got false
       /bin/sh -c stat\ -c\ \%a\ /etc/sudoers\ \|\ grep\ --\ \\\^288\\\$

     # ./spec/localhost/sudo_spec.rb:9:in `block (2 levels) in <top (required)>'

Finished in 0.05873 seconds (files took 0.48314 seconds to load)
4

1 回答 1

2

1) 通常os[:release]包含像'6.5'. 因此，进行比较的正确方法是：

if os[:release].to_i == 6

或者：

if os[:release].split('.')[0] == '6'

2）因为带有前导零的数字被Ruby从八进制（0440）转换为十进制（288），并且Serverspec（Specinfra）以十进制到字符串的格式将其传递给底层的grep命令。

于 2016-01-03T11:03:37.200 回答