Re: regular expression

On Tue, 2010-06-01 at 16:35 +0100, Richard Quadling wrote: > On 1 June 2010 16:33, Ashley Sheridan <ash@ashleysheridan.co.uk> wrote: > > > > On Tue, 2010-06-01 at 16:31 +0100, Richard Quadling wrote: > > > > $re1 = '/^[a-z]++$/i'; > > $re2 = '/^[a-z ]++$/i'; > > > > > > > > -- > > ----- > > Richard Quadling > > "Standing on the shoulders of some very clever giants!" > > EE : http://www.experts-exchange.com/M_248814.html > > EE4Free : > > http://www.experts-exchange.com/becomeAnExpert.jsp > > Zend Certified Engineer : > > http://zend.com/zce.php?c=ZEND002498&r=213474731 > > ZOPA : http://uk.zopa.com/member/RQuadling > > > > > > Why the double ++ in the expressions there? Surely one + would match the 1 or more > > characters that you need and the second one would just be surplus? > > > > Thanks, > > Ash > > http://www.ashleysheridan.co.uk > > > > > > ++ doesn't give back. As there is no need to do any backtracking, this > is supposed to be a slight optimization. > > > > -- > ----- > Richard Quadling > "Standing on the shoulders of some very clever giants!" > EE : http://www.experts-exchange.com/M_248814.html > EE4Free : http://www.experts-exchange.com/becomeAnExpert.jsp > Zend Certified Engineer : > http://zend.com/zce.php?c=ZEND002498&r=213474731 > ZOPA : http://uk.zopa.com/member/RQuadling Ah, I ought to have guessed as it's you! ;) I didn't know about that in regex's, I've learnt something new today! Thanks, Ash http://www.ashleysheridan.co.uk