NetNews Usenet Archive 1992 #31/NN_1992_31.iso/spool/comp/unix/question/15134

Text File | 1993-01-03 | 1.8 KB | 47 lines

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From: [e-mail] (Leif Jensen)
Subject: Re: searching a file
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References: <[e-mail]> <[e-mail]>
Date: Mon, 4 Jan 1993 01:55:30 GMT
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In article <[e-mail]> [e-mail] (Jean-Pierre Radley) writes:
>Here's a command to print all words in /usr/dict/words having 1, 2 or 3 "l"s:
>
></usr/dict/words tr '[A-Z]' '[a-z]' |
>sed -n '
>/^\([^l]*l[^l]*\)\1\{0,2\}$/p
>'

Here is a method using awk:

awk -Fl 'NF >= 2 && NF <= 4' < /usr/dict/words

Actually, I *think* your sed command is invalid.  (Your tr command
also needlessly includes '[' and ']'.) The man page says "A
one-character RE followed by \{m\}, \{m,\}, or \{m,n\} ...."  I do not
think \1 can be followed by \{0,2\}.  The \{\} construction is also not
present in all versions of sed.  Anyway, your RE wouldn't do quite
what you think.  '^\([^l]*l\)\1$' matches not lines with exactly two
l's, but only lines of the form alal, where "a" is any sequence of
characters not containing 'l' or newline and "l" is a literal 'l'.  A
regular expression like '^\([^l]*l[^l]*\)\1$' matches only lines of the
form "albalb".  The easiest way to do what you wanted using sed is to
use the command:

sed -n '/^[^l]*l[^l]*$/p
/^[^l]*l[^l]*l[^l]*$/p
/^[^l]*l[^l]*l[^l]*l[^l]*$/p' /usr/dict/words

Hope this helps,

--
[e-mail]
Leif Jensen