Hi,

At Wed, 8 May 2002 02:47:35 +0900,
Ian Macdonald wrote:
> > My current work-around is /([^x]|^)ox+o(?!x)/, but then I need to prepend
> > $1 to the replacement, e.g.
> 
> This will work, too, and doesn't require the $1:
> 
> /(?!x).ox+o(?!x)/

This doesn't match strings start with /ox+o/ (have no char
before 'o'), and replaces the char before 'o' too.

-- 
Nobu Nakada