Alessandro [AkiRoss] Re
7/31/2007 2:12:00 PM
Mh well, to me it seems a normal regex processing (i mean, it *should*
require only one instruction, since this pattern can be read with just
one regex, even if ruby doesn't allow it... but it would be really
bad).
Anyway well, splitting it there are different ways to do it - thanks
for your sudjestion.
But if ruby make it possible with one call, i'd prefer to use it.
Thanks!
On 7/31/07, Jano Svitok <jan.svitok@gmail.com> wrote:
> On 7/31/07, Alessandro Re <akirosspower@gmail.com> wrote:
> > Hi there!
> > I'm Alessandro from Italy and I started using ruby some days ago,
> > so... Hello, Community! :)
> >
> > Well, I was trying to match a pattern multiple times. I tried both
> > with normal match() and scan(), but i can't get the desired result.
> >
> > The subject string is something like:
> > "1a2bend" or "beg1a2b3c4dend"
> > more generally, it should match /^beg(\d\w)*end$/ : always a begin and
> > ending pattern, and a unspecified number of central pattern.
> > The problem is that the central pattern must be extracted for every
> > time it's encountered.
> > For example, trying with
> > "x1A2B3C4Dz".scan /^(x)(\d\w)*(z)$/
> > returns
> > [["x", "4D", "z"]]
> > while i need something like
> > [["x", "1A", "2B", "3C", "4D", "z"]]
> >
> > Why does ()* match just the last one? How can i get all the ()* that it matches?
> >
> > Probabily i'm doing something wrong, but can't understand where :>
> Try:
>
> if "x1A2B3C4Dz" =~ /^(x)((?:\d\w)*)(z)$/
> a, b = $1, $3 #
> return [a] + $2.scan(/\d\w/).flatten + [b]
> end
>
> I don't know if it's possible to do it in one run though, maybe you
> could use split as well...
> Take care when doing nested searches as they will overwrite $1..9
> (that's why I used a and b)
>
> J.
>
>
--
~Ale