Date: Tue, 07 Jul 2009 15:49:27 +0100 From: Matthew Seaman <m.seaman@infracaninophile.co.uk> To: "Aryeh M. Friedman" <aryeh.friedman@gmail.com> Cc: freebsd-questions@freebsd.org Subject: Re: ot: regular expression help Message-ID: <4A536077.1080002@infracaninophile.co.uk> In-Reply-To: <4A5353D1.5010807@gmail.com> References: <4A5353D1.5010807@gmail.com>
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Aryeh M. Friedman wrote:
> I am attempting to make (without the perl expansions) a regular=20
> expansion that when used as a delim will split words on any=20
> punction/whitespace character *EXCEPT* "$" (for java people I want to=20
> feed it into something like this:
>=20
> for(String foo:input.split([insert regex here])
> ...
Well, there's no way to say "all foo except bar" using standard regexes, =
so
you can't use the [:punct:] character class. You'll have to roll your own=
class.
If your input is ASCII then see ispunct(3) for a handy list of all the
ascii punctuation characters. I guess you'll need a RE something like th=
is:
[]!"#%&'\(\)\*\+,\./:;<=3D>?@[\\^_`{\|}~-[:space:]]+
although that's completely untried, quite likely to not have all the
metacharacters properly escaped (exactly what is or isn't a metacharacter=
depends on the RE implementation you're using) and is probably horribly
confused due to the inclusion of '[' '-' and ']' amongst the characters
matched in the range. =20
If you're using anything other than ascii, then I suspect you're going
to have problems with RE libs anyhow, unless you can somehow use PCRE. =20
The \p{isPunct} and \p{isWhite} escapes for matching unicode punctuation
or whitespace is probably what you need.
Even so, your best choice would probably be to separately check strings
for the presence of $ characters -- maybe transform those $ characters to=
something else -- and then split on any remaining punctuation characters.=
Cheers,
Matthew
--=20
Dr Matthew J Seaman MA, D.Phil. 7 Priory Courtyard
Flat 3
PGP: http://www.infracaninophile.co.uk/pgpkey Ramsgate
Kent, CT11 9PW
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