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