How to list users with UIDs higher than 10?

Question

What commands can I use to display all users who have an identifier (UID) that is greater than 10?

Solution 1

You can parse the passwd database using awk. In the passwd format, each entry is a list of fields delimited by :, with the first field being the username and the third field being the UID. So, in awk, you could do:

getent passwd | awk -F: '$3 > 10 {print $1}'

Solution 2

SOLUTION 1 :

You can use:

#!/bin/bash
while IFS= read -r line; do
    [[ "$(cut -d: -f3 <<<"$line")" -gt 10 ]] && echo "$line"
done </etc/passwd

Considering you have no username containing :.

If you just want the usernames:

#!/bin/bash
while IFS= read -r line; do
    [[ "$(cut -d: -f3 <<<"$line")" -gt 10 ]] && echo "$(cut -d: -f1 <<<"$line")"
done </etc/passwd

Here we are reading each line of the /etc/passwd file and comparing the third field delimited by : (UID) to check whether it is greater than 10. If it is true, we have printed the line (or username in second script).

This would be much simpler using an array (thanks to muru):

#!/bin/bash
while IFS=: read -a line; do
    [[ "${line[2]}" -gt 10 ]] && echo "${line[0]}"
done </etc/passwd

SOLUTION 2 :

Using grep with PCRE:

getent passwd | grep -P '^[^:]+:[^:]+:(?!(?:\d|10):)' | cut -d: -f1

• ^[^:]+: will match the first field (username) including trailing :

• [^:]+: will match the second field (password) including trailing :

• (?!(?:\d|10):) part is tricky, it is a zero width negative lookahead patern of PCRE (grep -P), it means there is no single digit (\d) number or number 10 next, followed by a :.

• cut -d: -f1 will just print the first field (username).

Alternately, you can avoid the cut, by using the negative lookahead (?!) inside of the positive lookahead (?=) pattern (thanks to Avinash Raj) :

getent passwd | grep -Po '^[^:]+(?=:[^:]+:(?!(?:\d|10):))'

SOLUTION 3 :

using python :

#!/usr/bin/env python2
with open('/etc/passwd') as f:
    print '\n'.join([line.split(':')[0] for line in f if int(line.split(':')[2]) > 10])

Here we are printing usernames if the third field is grater than 10. line.split(':') will split fields of each line on :, making it a list of fields.

Solution 3

Using awk:

 awk -F: '{if ($3 > 10) { print $1 ":" $3 } }' /etc/passwd

this will list all users with their associated UID where UID > 10.

Thanks to @sadi note, to list only usernames

awk -F: '{if ($3 > 10) {print $1}}' /etc/passwd

Comments

[Panther]

You may simply awk -F ':' '$3 > 10' /etc/passwd . print is the default behavior, less typing

[muru]

@bodhi.zazen as a frequent user of LDAP systems, I see no reason not to use getent. And I suppose OP only wants the username.

[muru]

Set IFS=:, and use read -a, then you can use ${line[0]} and ${line[2]} directly.

[heemayl]

@muru thank you very much..i have not thought it that way..adding..

[Avinash Raj]

@heemayl i think you may try ^[^:]+(?=:[^:]+:(?!(?:\d|10):)), you don't need for cut command.

[Avinash Raj]

through python, with open(file) as f: for line in f: m = line.split(':'); if int(m[2]) > 10: print(m[0]) . Note that this is not an one-liner.

[Sadi]

This lists all usernames followed by a column and UID. However this lists usernames only: awk -F: '{if ($3 > 10) {print $1}}' /etc/passwd

[Maythux]

@Sadi thanks but I insert the UID to show each User with its associated UID

[Sadi]

I understand; I just wanted to add a note how to modify this in case only usernames are required - UpVoted ;-)

[heemayl]

@AvinashRaj Thank you very much..neat..added..