Date: Thu, 13 Jan 2011 18:22:18 -0600 (CST) From: Robert Bonomi <bonomi@mail.r-bonomi.com> To: freebsd-questions@freebsd.org, freebsd@edvax.de Subject: Re: awk question: replacing "%d%s" by "%d %s" Message-ID: <201101140022.p0E0MIW8029158@mail.r-bonomi.com> In-Reply-To: <20110113062819.4ecb89d9.freebsd@edvax.de>
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> Date: Thu, 13 Jan 2011 06:28:19 +0100
> From: Polytropon <freebsd@edvax.de>
> Subject: awk question: replacing "%d%s" by "%d %s"
>
> I'm aware that this is not an awk question list, but I'm confident there
> are many awk gurus here who can surely help me with such a stupid
> problem. I also know that I get more and more stupid myself for NOT being
> able to solve this, even after... some nearly infinite time. :-)
>
> I have strings of the form either "<number(s)>" or
> "<number(s)><letter>". I catch them with
>
> if(match(nr, "[a-z]"))
> ...
>
> where "nr" is the name of the string. What I need is a simple space
> between <number(s)> and <letter>, so for example "12a" would get "12 a",
> "6d" would get "6 d", and "58" would stay unchanged. I've tried with
> split(), with array manipulation and could produce 10 lines of code that
> didn't work as intended
> (it produced "1122aa", "66dd" and "5588" according
> to the examples above).
> Obviously, sub(nr, "[a-z]", " [a-z]"); is nonsense.
>
True. But
sub(nr,"[a-z]"," &");
does the trick. (tested on Freebsd 7.2)
Explamation: "&" is a 'replacement side' magic incantation to the regex
library that means 'that which was matched by the pattern regex'.
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