Ubuntu Tutorials

sed -i ‘/^$/ d’ usernames.txt

I came across this little introduction lately. as a non geek it reads … intresting…
http://phi.lho.free.fr/programming/RETutorial.en.html

egrep “\w” usernames.txt

grep . usernames.txt

If the file has a chance of duplicates, I’d actually do:

sort -u usernames.txt |grep .

VIM:
:g/^\s*$/d

[dcarter@host ~]$ grep . usernames.txt
jstark
bchilds
cedwards
arhodes
mrowley
prhodes
asmith
bcorey
hnakamura
ppetrelli
cbyrd
jstrazzo
sraver
bgates
ltorvalds

grep -v ‘^$’

cat usernames.txt | sort | uniq | tail -n `cat usernames.txt | wc -w`

just because I hate regexes and avoid them if I can.

get-content C:\Scripts\usernames.txt | foreach {if ($_.length -ne 0) {write-host $_}}

xargs printf “%s\n” < usernames.txt

(and this week’s Useless Use of Cat Award goes to…)

ruby -pe ‘next if $_=~/^$/’ < usernames.txt

Ok, now suppose the usernames aren’t all on separate lines. Sometimes there are multiple usernames on one line, comma separated, maybe there’s a space after the comma, but maybe there’s not. Like this:

jstark

bchilds, cedwards,arhodes

mrowley

prhodes

asmith
bcorey

hnakamura
ppetrelli
cbyrd

jstrazzo
sraver
bgates

ltorvalds

Oh, my answer for the one I just suggested is at http://student.seas.gwu.edu/~mac/files/regex.html

Python:

for x in re.sub(“\n+”, “:”, file(“username.txt”).read().strip()).split(‘:’):
print x

sed ‘/^\s*$/d’ usernames.txt

perl -ne ‘print unless (/^\s*$/)’ usernames.txt

perl -ne ‘print if (/\S/)’ usernames.txt

egrep -v ‘^\s*$’ usernames.txt

awk ‘!/^\s*$/ { print }’ usenames.txt

vim -c ‘sil g/^\s*$/d’ usernames.txt

Although it’s aimed at sed I’ve always found this a useful source of regex solutions : http://sed.sourceforge.net/sed1line.txt

Mackenzie:
sed -e ‘s/,/\n/g’ -e ‘s/[ \t]//g’ -e ‘/^$/d’ usernames.txt

Converts commas to newlines, strips whitespace, then removes blanks.

Mark.

Regex can be avoided in most cases and should be when possible.

Try this:
awk ‘/^[a-zA-Z]/{print $1}’ usernames.txt

Unlike just removing the newlines, that prevents cases where some crazy admin has commented out users or whatnot in /etc/passwd.

If that was a standard passwd file you could do something more like:
awk -F: ‘/^[a-zA-Z]/{if ($3 !~ 0) print $1}’ /etc/passwd

If it is and Ubuntu / Debian server where normal uids start at 1000, you could do this for a listing of usernames:
awk -F: ‘/^[a-zA-Z]/{if ($3 >= 1000 && $1 != “nobody”) print $1}’ /etc/passwd

Usernames must start with a character and can’t start with a number from my understanding so that should work fairly well.

Just delete blank lines:
perl -pi -e ‘s/\A\s+\z//’ usernames.txt
With commas too:
perl -pi -e ‘s/,\s*/\n/g;s/\A\s+\z//’ usernames.txt

import sys

for line in sys.stdin:
usernames = [name.strip() for name in line.split(",")]
if not any(usernames):
continue
for username in usernames:
print username

Readability – 1
The way of the Pathologically Eclectic Rubbish Lister – 0

#!/usr/bin/perl
open(FILE,”usernames.txt”) || die “Could not open usernames.txt: $!\n”;
foreach () {if ($_ =~ /\w+/) {print “$_”}}
close(FILE);

Of course you’d have to be crazy to use perl for something that easy in grep/awk!!

boo that should be foreach \…

Hm… why do I have to post this three times? :-/

awk -v RS=”[\n ,]+” ‘{print}’ < u.txt

Late to the challenge, but here’s the regex I whipped out off the top of my head:
s/\n\W*\n/\n/g

My perl code slurps the entire file to a variable and regexes that variable, then splits on newline to an array.

emacs22

C-M-%

C-q C-j C-q C-j +

C-q C-j

and it will then ask you “[y/n]?”

I second “grep .” as the easiest solution.

Perl/Ruby shouldn’t be too complicated either:

perl -lane ‘print if /./’
ruby -lane ‘p if /./’

Too filter whitespace lines (not only blanks):

perl -lane ‘print if /\S/’
ruby -lane ‘print if /\S/’