From: Liu Lantao <liulantao@...>
Date: 2009-12-01T22:47:34+09:00
Subject: Re: one more regular expression, please

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extractor = Regexp.new('\+\+\(\((\d+)\)\)\+\+')
extractor.match will extract all the ++(123)++ like parts and extract all
123s into an array,
if the string start with ++, try
^\+\+\(\((\d+)\)\)\+\+

On Tue, Dec 1, 2009 at 10:07 AM, J. mp <joaomiguel.pereira@gmail.com> wrote:

> Hi,
>
> I really can't figure out how regular expressions work, and since I
> don't use them frequently is easier to ask...
>
> I have the follwow string
>
> ++((100))++askjasdjksjdkasdkjk++((200))++skdjsjd
>
> I need a reg expression that extracts only the first number inside the
> first ++(( ))++ and ignore the rest..., in this case I need to return
> 100
>
>
> thanks
> --
> Posted via http://www.ruby-forum.com/.
>
>


-- 
Liu Lantao
College of Information Science and Technology, Beijing Normal University
EMAIL: liulantao ( at ) gmail ( dot ) com ;
WEBSITE: http://www.liulantao.com/ .
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