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
