David -- ...and then David Newman said... % % > Yes, but your *. should be .* to work. When you have % % Sorry, typo. It is .* in the code!
Ah. Well, then, there's more... % ... % > % Again, my goal is to match each instance of /(p1)*anything*(p2)/. % > % > I should think that % > % > undef $/ ; % > ... % > /(p1).*(p2)/g ; % > % > would do it for you. Whoops! No it won't -- .* is greedy! You need something more like undef $/ ; ... /(p1)[^(p2)]*(p2)/g ; in order to match p1 and the *first* p2 each time :-) Silly me... Back to the Camel book for *me*! HTH & HAND :-D -- David T-G * It's easier to fight for one's principles (play) [EMAIL PROTECTED] * than to live up to them. -- fortune cookie (work) [EMAIL PROTECTED] http://www.justpickone.org/davidtg/ Shpx gur Pbzzhavpngvbaf Qrprapl Npg!
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