Re: bash, grep, and regular expressions

[Justin Guerin <jguerin@cso.atmel.com>]

On Thursday 17 February 2005 17:08, Freddy Freeloader wrote:
> Hi All,
>
> I'm pretty much a noob at regular expressions but have been studying
> them lately using Jeffrey Friedl's book Mastering Regular Expressions
> 2nd ed.. I've run across something that has me puzzled. Here's what it
> is.
>
> Now before you flame me this is done strictly as an exercise in regular
> expressions.
>
> I have ls aliased to ls -al.  What I've been attempting to do with grep
> and regular expressions is list only non-hidden directories and/or
> files.  I am unable to come up with an expression that will elimate
> hidden files and return non-hidden files at the same time.
>
> ls -al | grep -v ' \.\<[a-zA-Z0-9].*\>' # returns everything
>
> ls | grep -e '\<[^.][[:alnum:]]'  # returns everything
>
> ls | grep -e '\<[.][[:alnum:]]'  # returns an empty set
>
>
> I've tried many more expressions than these but none of them have worked
> so far and the first one works on a Solaris 9 machine with a bash shell.
>
> I also cannot write a back reference that uses parenthesis.  I copied
> examples right out of Friedl's book and they won't work either.  I get a
> "invalid back reference" error every time I try.
>
> Anyone have any insight into this?  I'm running sarge with a 2.6.9
> kernel on one Debian box and a 2.6.10 kernel on another, and all
> packages are the very latest packages available in sarge on both boxes.
>
> TIA
>
> Gary

Hi Gary,

How about ls -la | grep -v -w "\..*" ?  Works here:
uname -a: Linux jguerin 2.6.6-1-686 #1 Wed May 12 14:57:57 EST 2004 i686 
GNU/Linux
bash --version
GNU bash, version 3.00.14(1)-release (i386-pc-linux-gnu)
Copyright (C) 2004 Free Software Foundation, Inc.

>From the grep man page:
-w, --word-regexp
 Select only those lines containing matches that form whole words.  The test 
is  that  the  matching  sub-string  must  either  be  at  the beginning of 
the line, or preceded by a non-word constituent character.
 Similarly, it must be either at the end of the line or followed  by  a  
non-word  constituent  character.
 Word-constituent characters are letters, digits, and the underscore.

The problem with just using grep -v "\..*" is that the . can appear 
anywhere.  The problem with grep -v "^\..*" is that the . can only appear 
at the beginning of a line.

Let me know if this works for you.

Justin Guerin