I've had a good look round the FAQs, but I can't work this one out.

I want to find every occurrence of ABC _not_ followed by XX\d+ (e.g.
ABCXX123 is fine, ABCXY123 is not).  I tried using this:

/ABC[^(XX\d+)]/

which doesn't work - I guess everything within [] is just being treated
as a character class, so the parentheses and + lose their meta-ness.

I can see how to solve this programmaticlly (i.e. search for ABC, then
check XX\d+ for each one found), but this would mean I'd be looping in
Perl for every _valid_ use of ABC, whereas I only want to process
"errors".

Does anyone know how to do it in a single expression?

Thanks,
John Edwards

On Tue, 23 Sep 1997 11:58:54 +0100,

--

Is that what you want?

--
Tom Phoenix           http://www.teleport.com/~rootbeer/

Randal Schwartz Case:  http://www.rahul.net/jeffrey/ovs/
              Ask me about Perl trainings!

well... this should work:
/ABC(?!XX\d+)/

The (?!...) is the negative look-ahead operator... it means: when whatever
is NOT followed by whatever is here in parens ().

The opposite is (?=...).  it means: when whatever IS followed by whatever
is here in parens ()...

check out "man perlre" for more on this.

----------------
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----------------



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   I've had a good look round the FAQs, but I can't work this one out.

   I want to find every occurrence of ABC _not_ followed by XX\d+ (e.g.
   ABCXX123 is fine, ABCXY123 is not).  I tried using this:

[snip]

   Does anyone know how to do it in a single expression?

(?! ) specifies a zero-width negative lookahead, e.g.

/ABC(?!XX\d+)/

-Jon

------------------------------------
Jon Orwant            http://tpj.com
Editor & Publisher, The Perl Journal