From: Andrew Johnson <ajohnson@...>
Date: 2006-01-25T16:13:13+09:00
Subject: Re: subsequence regular expression

On Wed, 25 Jan 2006 15:07:31 +0900, diz rael <dizraelus@yahoo.com> wrote:
[snip] 

> Anyway, I tried the following:
> 
> s=3D~/.+?(ba)+/
> $& =3D> "5b300ba"
> 
> It so happens that a "ba" was created on the stack, so
> the regexp thinks it ends there, when in fact the
> string I want it to return is:
> 5b300ba00260


Well, you could just check for 2 or more 'ba' sequences:

  s =~ /(.*?)(ba){2,}/
  p $1

but of course, your "dirty portion" just might just have two, or even more,
stray 'ba' in a row -- so I suspect you want to grab everything from the
beginning up to the longest subsequence of repeated 'ba's? One way:

  s = "5bbaba300ba00260babababababababababababa000bd1007b810ba92"
  puts s[0,s.index(s.scan(/(?:ba)+/).max)]

But I may well be missing an easier way :-)

cheers,
andrew

-- 
Andrew L. Johnson   http://www.siaris.net/
      The generation of random numbers is too 
      important to be left to chance.