如何在不以“特定字符”结尾的 bash 中显示所有用户

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.