![引用标题数组中的每个标题](https://linux22.com/image/615758/%E5%BC%95%E7%94%A8%E6%A0%87%E9%A2%98%E6%95%B0%E7%BB%84%E4%B8%AD%E7%9A%84%E6%AF%8F%E4%B8%AA%E6%A0%87%E9%A2%98.png)
使用 Puppet 2.7.11,我需要创建从/usr/local/bin
到的几个符号链接/usr/bin
,并且想要巧妙一点:
class containing_class {
file { [ "/usr/local/bin/job", "/usr/local/bin/jstart",
"/usr/local/bin/jstop", "/usr/local/bin/jsub"]:
ensure => link,
target => regsubst(name, "^/usr/local/bin/", "/usr/bin/")
}
但是,name
会导致链接的目标为name
,$name
并$title
使用containing_class
(!)代替,并且$path
值为$PATH
。
如何在通话中引用单个文件的标题/路径?
答案1
嗯,这可能不是你想要的,但应该可行:
class containing_class {
define bin_link {
file { $title:
ensure => link,
target => regsubst($title, "^/usr/local/bin/", "/usr/bin/"),
}
}
bin_link { [ "/usr/local/bin/job", "/usr/local/bin/jstart",
"/usr/local/bin/jstop", "/usr/local/bin/jsub"]: }
}