${name} returns class name instead of resource namevar

[Francis Pereira]

When using a custom resource to deploy java $name is not the name of the resource but the name of the class in the below mentioned example

 java::setup { "${myapplication::params::app_username}-jdk_7u51":

.......

    cachedir      => "/var/lib/puppet/file_cache/${name}-working",

  }

Shouldn't $name = ${myapplication::params::app_username}-jdk_7u51 ?

Francis Pereira

[José Luis Ledesma]

No, it is the current $name, not the resource one.

Regards,

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[jcbollinger]

No.  The situation is analogous to a function call: if the argument list contains a variable, the value passed is the value of the variable in the calling scope, not the value of any same-named local variable in the called function.


John

[Johan De Wit]

Hi,

I think you are mixing the 'define' a  resource  and 'use' resource type.

$name is an automatic variable that is set by puppet, and will be in the case of a defined resource, the resource title you give it when 'using' that defined resource

You can use the $name in the define <your type> { } , and in that scope it will be given your example :  ${myapplication::params::app_username}-jdk_7u51.

Outside the definition of your resource (the file beginning with define <whatever> {} ) the $name var contains the $name of the class you are using your defined resource .

I hope i did not confuse you more now.

but read http://docs.puppetlabs.com/puppet/latest/reference/lang_defined_types.html#title-and-name

especially the part about uniqueness. 

An to make it much more difficult, one should start using $title in defined resources instead of $name.

the why is explained in the same link I gave.

I hope it helps.

Grts
Jo

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Johan De Wit

Open Source Consultant

Red Hat Certified Engineer              (805008667232363)
Puppet Certified Professional 2013/2014 (PCP0000006)
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