For loop and echo in unix

Question

I am trying to write a shell script to list all the users who have id 101

 #!/bin/bash who="$(who | cut -d ' ' -f 1 | sort -u )"; #Save the output of who for user in ${who}; do # Iterate over $@ if [ $(echo id -g $user) == "101" ] ; then echo "Got it"; fi done; 

when i execute my script I get this error "line 4: [: too many arguments". I am not sure where I made mistake.

Answer 1

Why not look directly at /etc/group?

awk -F: '$3==101 { print $4 }' 

If getent is available on your host, you can do this to get a list pulled from /etc/passwd which is then chewed on:

awf -F: '{print $1}' <(getent group 101) 

Answer 2

The error comes from the fact that $( echo id -g $user ) will be expanded into the words id -g username. This can not be compared to 101 since the expansion is unquoted.

To compare the output of id -g "$user" (note the double quotes), use

if [ "$( id -g "$user" )" = "101" ] 

Within [ ... ] you should use a single = to do string comparison. In shells that have [[ ... ]] you may use ==:

if [[ $( id -g "$user" ) == "101" ]] 

Here, the quoting of the command substitution is not required, but it is if you use [ ... ].

---

The idiomatic way to do this sort of task is not to store the output of the who pipeline in a variable, but to pass it directly to the loop:

who | awk '{ print $1 }' | sort -u | while read user; do if [ "$( id -g "$user" )" = "101" ]; then echo 'Got it' fi done 

Answer 3

I believe changing the line to be:

if [ $(id -g $user) == "101" ]; then 

will solve the issue. The original command had an echo in the output, which I believe is not needed.

Answer 4

Use getent to query the password database:

getent passwd 101