From: Peter Pk <peterk@...>
Date: 2009-05-20T03:41:26+09:00
Subject: Re: rake namespace selection

Peter Pk wrote:
> 
> is there a way to "explicitly" invoke a block inside the scope of a
> parent or explicit "root relative" rake namespace instead of always
> being a child of the current namespace ???
> 
> 
> namespace :xxx do
>   namespace :yyy do
>     namespace ...
> 
>           ... called_in_unknown_namespace(arg);
> 
>     end
>   end
> end
> 
> 
> def called_in_unknown_namespace(arg)
> 
>   # execute below block as if it were in "root" namespace
>   namespace 'root namespace' do
> 
>     # declare additional item to do in root namespace task
>     task :task_in_root do
>       puts('invoked by "rake task_in_root"');
>     end
>   end
>   # scope restored to wherever you were previously
> 
> end
> 
> Thanks!

Ok - figured it out looking at the source, since all tasks simply have 
the "scope name" prepended to their name, and the "namespace" is just a 
list of names that defines the currnet scope scope in the task manager, 
then I was able to do the below to explicitly selct the "global scope". 
building the list from a scope name ie: ':name1:name2:name3', '^name' 
using the same format as a task name reference will make it general. 
BTW what would be appropriate to specify an "absolute path" in rake 
convention ???? since namespace :xxx specifies a relative *child* - ie: 
desirably somethign like file paths '/' and '..' maybe '::' ???

module Rake
  module TaskManager
    def in_explicit_namespace(name)
      oldscope = @scope;
      @scope = Array.new();
      # build scope name list from name here
      ns = NameSpace.new(self,@scope);
      yield(ns)
      ns
    ensure
      @scope = oldscope;
    end
  end
end

# down below in a rakefile

namespace :num1 do
  namespace :num2 do
    Rake.application.in_explicit_namespace ':' do
      puts("in explicit block");

      task :x do |t|
        puts(" doing explicit task x");
      end
    end
  end
end

# from command line

$ rake x
  doing explicit task x
$





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