Puppet：“$x.each...”出现语法错误

Question

我有以下代码：

define keepalived::vrrp_instance(
  $state,
  $interface,
  $virtual_addresses,
  $virtual_router_id,
  $priority      = $::keepalived::params::priority,
  $advert_int    = $::keepalived::params::advert_int,
  $password      = $::keepalived::params::password,
  $notify_master = $::keepalived::params::notify_master,
  $notify_backup = $::keepalived::params::notify_backup,
  $notify_fault  = $::keepalived::params::notify_fault,
  $notify_all    = $::keepalived::params::notify_all,
  $smtp_alert    = $::keepalived::params::smtp_alert,
) {
  ...
  $virtual_addresses.each |$address| {
    $splitted_address = split($address,' ')
    if  !is_ip_address($splitted_address[0]) {
      fail("Error virtual_address Value: \"${address}\" not an ip address!")
    }
  }
  ...
}

$virtual_addresses 类似于 ['127.0.0.1 dev eth0','fd00::1 dev eth0']

运行代码我得到以下错误：

Syntax error at '.'; expected '}' at /etc/puppet/environments/ip6_dev/modules_custom/keepalived/manifests/vrrp_instance.pp:136 on node

第 136 行是“$virtual_addresses.each |$address| {”

我找不到错误(https://docs.puppetlabs.com/references/3.stable/function.html#each)

我正在使用 Puppet 3.3.2

Answer 1

"note requires parser = future"

确保您在 Puppet 中使用未来的解析器。在 puppet.conf 文件中设置 parser = future 或添加命令行开关 --parser=future

更新： 包装您的验证功能：

define verify::wrapper ()
{
    $ip_address = split($name,' ')
    if  !is_ip_address("${ip_address[0]}") {
        fail("Error virtual_address Value:  \"${ip_address[0]}\" not an ip address!")
    }
}

下次使用它：

define keepalived::vrrp_instance(...)
{
    ...
    verify::wrapper{ $virtual_addresses : }
    ...
}

Puppet：“$x.each...”出现语法错误

Puppet: Syntax error with '$x.each...'

each

syntax-error

puppet