Date: Fri, 14 Jan 2011 08:47:11 +0100 From: Polytropon <freebsd@edvax.de> To: ws@au.dyndns.ws Cc: freebsd-questions@freebsd.org, Robert Bonomi <bonomi@mail.r-bonomi.com> Subject: Re: awk question: replacing "%d%s" by "%d %s" Message-ID: <20110114084711.0e76b8c2.freebsd@edvax.de> In-Reply-To: <1294989784.2037.79.camel@predator-ii.buffyverse> References: <20110113062819.4ecb89d9.freebsd@edvax.de> <201101140022.p0E0MIW8029158@mail.r-bonomi.com> <20110114071712.c7a7fe7e.freebsd@edvax.de> <1294989784.2037.79.camel@predator-ii.buffyverse>
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On Fri, 14 Jan 2011 17:53:04 +1030, Wayne Sierke <ws@au.dyndns.ws> wrote:
> I suspect it is a transcription error by Robert in his email.
>
> From man awk:
>
> sub(r, t, s)
> substitutes t for the first occurrence of the regular
> expression
> r in the string s. If s is not given, $0 is used.
>
>
> So the correct syntax is:
>
> sub("[a-z]", " &", nr)
Works in this version, thanks! Reduction of 7 lines of code.
sub(from, to, where); is the correct form.
--
Polytropon
Magdeburg, Germany
Happy FreeBSD user since 4.0
Andra moi ennepe, Mousa, ...
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