From: dblack@...
Date: 2002-08-29T05:48:52+09:00
Subject: Re: Newbie converting brain from perl

Hello --

On Thu, 29 Aug 2002, William Pietri wrote:

>
> Howdy! As a long time Perl, Java, and Objective-C developer, I'm quite taken
> with Ruby, and I'm busily converting my brain over. But there's a Perl
> idiom that I'm having trouble translating. In Perl it goes like this:
>
>     $date = "28 Aug 2002";
>     ($day, $month, $year) = $date =~ /(\d+) (\w+) (\d+)/;
>
>
> In Ruby the best I've been able to do is this:
>
>     date = "28 Aug 2002"
>     day, month, year = /(\d+) (\w+) (\d+)/.match(date)[1..3]
>
> But that final [1..3] thing seems inelegant, which makes me think I'm
> barking up the wrong tree. What's the best Ruby-style way to do this?

Well, my response to this problem was to participate in a coding
retreat where we implemented scanf :-)  (See earlier thread on this.)

  require 'scanf'

  d = "28 August 2002"
  day,month,year = d.scanf("%d%s%d")
  p day, month, year

  =>  28
      "August"
      2002

(You can get scanf at <http://www.rubyhacker.com/code/scanf>.)

You could also do:

  day,month,year = d.scan(/(\d+) (\w+) (\d+)/).flatten

You need to flatten the scan results because scan returns an array of
()-captured substrings for every successful scan through the array --
so otherwise, day will be a three-element array and month and year
will be nil.

(Note that scanf gives you the %d fields as numbers (in case that's a
consideration), whereas match and scan give you strings.)


David

-- 
David Alan Black
home: dblack@candle.superlink.net
work: blackdav@shu.edu
Web:  http://pirate.shu.edu/~blackdav