asked on

Linux - members in group less than 99

Member of groups which have gid less than 99 ( root or any ) , I need a command that will give me output of the usere which are in this groups

Please help me with the command

BlackAdderDK

Hi

The following script should do the trick:

#!/bin/bash
let userid=100

IFS="
"
for i in `cat /etc/passwd`
do
let j=`echo $i | cut -f 3 -d ':' 2>/dev/null`

if [ $j -le $userid ]
then
echo $i|awk -F":" '{ print "Username: " $1 "\t\tUID:" $3 }'
fi
done

Regards
'Adder

Brad_Wosmek

Assuming all users are in /etc/passwd you could use this grep:

grep -E .*\:.*\:.*\:$[0-9][0-9]\|[0-9]$\:.* /etc/passwd | cut -d: -f1

This expression will include gids <=99 not <99. I suspect this is what you want though.

BlackAdderDK

Hi... again

Maybe I should read the whole ;)

This fits :)

#!/bin/bash
IFS="
"
for i in `cat /etc/passwd`
do
let j=`echo $i | cut -f 4 -d ':' 2>/dev/null`

if [ $j -le "99" ]; then
echo $i|awk -F":" '{ print "Username: " $1 "\t\tGID:" $4 }'
fi
done

regards
'Adder

singhjagsir

Just one beautification to output, Idea/Script is same as BlackAdder

#!/bin/bash
IFS="
"
for i in `cat /etc/passwd`
do
let j=`echo $i | cut -f 4 -d ':' 2>/dev/null`

if [ $j -le "99" ]; then
echo $i|awk -F":" '{ print "GID: " $4 "\t\tUsername: " $1 }'
fi
done |
sort -n -k2

_J

#!/bin/bash
IFS="
"
for i in `cat /etc/passwd`
do
let j=`echo $i | cut -f 4 -d ':' 2>/dev/null`

if [ $j -le "99" ]; then
   echo $i|awk -F":" '{ print "GID: " $4 "\t\tUsername: " $1 }'
fi
done |
 sort -n -k2

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simon3270

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