From: Vincent Fourmond <vincent.fourmond@9online.fr>
Date: 2007-02-27T21:31:58+09:00
Subject: Re: Regular expression matches last occurrence instead of first

andyo wrote:
> Here's the sample code; note that (.*?) and ([^"]+) behave the same
> way--and not the way I'd expect:
> 
>     str = '"aaaaa""bbb""ccc"'
> 
>     str.scan(/"(.*?)"/)
>     puts $1
>     # ccc

  Normal... #scan is not what you'r looking for:

------------------------------------------------------------ String#scan
     str.scan(pattern)                         => array
     str.scan(pattern) {|match, ...| block }   => str
------------------------------------------------------------------------
     Both forms iterate through str, matching the pattern (which may be
     a Regexp or a String). For each match, a result is generated and
     either added to the result array or passed to the block. [...]

  scan find all successive matches for the pattern, and sets the
captured groups variables everytime it finds one. So, here, you simply
get the $1 for the last match, ie "ccc".

  What you're looking for is simply =~, as in Perl:

irb(main):001:0> str = '"aaaaa""bbb""ccc"'
=> "\"aaaaa\"\"bbb\"\"ccc\""
irb(main):002:0> str =~ /"(.*?)"/
=> 0
irb(main):003:0> $1
=> "aaaaa"

  Cheers,

	Vincent
-- 
Vincent Fourmond, PhD student (not for long anymore)
http://vincent.fourmond.neuf.fr/