"z u i u" =~ /(z|u|\s)+?/
will match "z".
The "?" enforces a "minimal" match (whereas by default a "greedy" match is
done)!
Regards
Clemens
> -----Ursprgliche Nachricht-----
> Von: Dave Thomas [mailto:Dave / PragmaticProgrammer.com]
> Gesendet am: Dienstag, 7. August 2001 14:50
> An: ruby-talk / ruby-lang.org
> Betreff: [ruby-talk:19288] Re: re
> 
> stesch / no-spoon.de (Stefan Scholl) writes:
> 
> > irb(main):001:0> "z u i u" =~ /(z|u|\s)+/
> > 0
> > 
> > First match on position 0. But $1 is " " instead of "z".
> 
> The regexp matches "z u ", so it starts at position 0.
> 
> The last parenthesized expression matched the final space.
> 
> 
> 
> Dave
>