CODEWITHSUNDEEP
linuxbash
Rafael Monteiro
Rafael Monteiro

Reputation: 13

How to show all users in bash which does not end with "specific character"s

All the users in the system that dosen't have as an ending character on their names a, s, t, r, m, z must be shown in Bash.

The users names can be obtained from the /etc/passwd file, in the first column. But I can not perceive the correct approach to exclude those characters from the search.

Should I use grep? Or just a cut?

Upvotes: 1

Views: 399

Answers (3)

Shawn
Shawn

Reputation: 52529

Something like

grep -o '^[^:]*[^astrmz:]:' /etc/passwd | tr -d :

or

cut -d: -f1 /etc/passwd | grep '[^astrmz]$'

[^blah] matches any character but the ones listed, the opposite of [blah].

GNU grep using a lookahead:

grep -Po '^[^:]*[^astrmz:](?=:)' /etc/passwd

Or using awk instead:

awk -F: '$1 ~ /[^astrmz]$/ { print $1 }' /etc/passwd

Or in pure bash without external commands:

while IFS=: read -r name rest; do
  if [[ $name =~ [^astrmz]$ ]]; then
    echo "$name"
  fi
done < /etc/passwd

As you can see, there's lots of potential approaches.

Upvotes: 3

rtx13
rtx13

Reputation: 2610

Simple one liner when using bash:

compgen -u | grep -v '[astrmz]$'

The compgen -u command will produce a list of users (without all of the extra fields present in /etc/passwd); compgen is a builtin in bash where it's normally used for username completion.

Upvotes: 1

Mamun
Mamun

Reputation: 2534

This should do the trick:

cut -d: -f1 /etc/passwd | grep -vE 's$|t$|r$|m$|z$'

The cut command strips out the username from the password file. Then grep -v (does the UNMATCHING) grep -E does multiple matching (OR OR OR) the $ sign indicates the last character to match

For example , on my mac, I get:

_gamecontrollerd
_ondemand
_wwwproxy
_findmydevice
_ctkd
_applepay
_hidd
_analyticsd
_fpsd
_timed
_reportmemoryexception

(You see no names end with those 5 letters).

Good Luck.

Upvotes: 0

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