From: James Britt <james_b@...>
Date: 2006-05-14T03:40:59+09:00
Subject: Re: list folder content from win directory -> xml output

Sergey Volkov wrote:
> just to continue after previous start :) :
> #
> open( "all_files.xml", "w" ){ |out|
>    out.puts "<all_files>"
>    Dir["**/*"].each{ |f| out.puts "  <file>#{f}</file>" }
>    out.puts "</all_files>"
> }
> #
> this code outputs list of all files in output file in xml format: 
> 'all_files.xml';

You may need to replace special characters

class String
   # For demo purposes only.
   def xscape
     self.gsub( '&', '&amp;').gsub( '<', '&lt; )
   end
end


And then call

     Dir["**/*"].each{ |f| out.puts "  <file>#{f.xscape}</file>" }


For extra fun consider adding file attributes as element attributes:

     Dir["**/*"].each{ |f|
       fl = File.new(f)
       mtime = " mtime='#{fl.mtime.to_s.xscape}' "
       ctime = " ctime='#{fl.ctime.to_s.xscape}' "
       out.puts "  <file #{mtime} #{ctime} >#{f.xscape}</file>"
      }

More or less.
> 
> resources: you can start with
> http://www.rubycentral.com/book/
> http://www.ruby-lang.org/en/

and

   http://www.ruby-doc.org

There may be some useful stuff at

   http://www.rubyxml.com

as well.


-- 
James Britt

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