how to display users under a group in a line for each user in Linux bash

Question

Currently when I run a grep command on a visitors group that I had created

grep visitors:x:1011: /etc/group

This is a sample result of the above command, as the actual result is too long.

visitors:x:1011:ellipsiscoterie,magnetcommonest,wizardmeans,beakskeletal,lemonwellmade,ralphpaperclips etc..

How do I display the above result as shown below (without visitors:x:1101 and separating the commas), as I have difficulty doing so.

ellipsiscoterie
magnetcommonest
wizardmeans
beakskeletal
lemonwellmade
ralphpaperclips

Answer 1

I suggest using sed instead of grep

sed -n -e 's/^visitors\:x\:1011\://p' /etc/group | tr ',' '\n'

UPD: detailed explain of sed part is here. tr examples you can find here

Answer 2

Using awk, you can set the field separated to a colon using -F: and check if the string starts with visitors:x:1011:

As the comma separated values always seems to be at the end, split the last field $NF on a comma and loop the splitted values.

awk -F: '
/^visitors:x:1011:/ {
  count=split($NF, ary, ",");
  for(i=1; i<=count; i++) print ary[i];
}' /etc/group

Answer 3

With your shown samples please try following awk program.

awk -F':' '/^visitors:x:1011/ && NF==4{gsub(/,/,ORS,$4);print $4}' /etc/group

OR in case entry visitors:x:1011 comes only once in your file then try following:

awk -F':' '/^visitors:x:1011/ && NF==4{gsub(/,/,ORS,$4);print $4;exit}' /etc/group

Explanation: Simple explanation would be, making field separator as : for all lines. Then in main program, checking condition if line starts from visitors:x:1011 and number of fields are 4 then globally substitute comma with new line(ORS) in 4th field and print it.

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OR if line with visitors:x:1011 comes once and 4th field is NULL(means no entry present which is obvious in some cases) and you want to print message for it then try following:

awk -F':' '/^visitors:x:1011/ && NF==4{if($4==""){print "NO entry found in /etc/group"} else{gsub(/,/,ORS,$4);print $4};exit}' /etc/group

Answer 4

A good approach to query groups and users is making use of the getent command as it also works on ldap systems.

$ getent group visitors | cut -d: -f4 | tr ',' '\n'

On standard systems this will always work as usernames and groupnames cannot contain the <comma>-character or the <colon>-character. (If you have a non-standard Linux or Unix system, see man useradd and man groupadd to check the username and group name rules)

If you want to use awk instead of this little pipeline, then an elegant—but cryptic—version would read:

$ getent group visitors | awk -F: '{FS=",";OFS=ORS;$0=$4;$1=$1}1'

Answer 5

One option could be to use awk:

awk -F: '$1 == "visitors" {
    size = split($4, users, ",");
    for(i=1; i<=size; ++i) print users[i];
    exit }' /etc/group

• -F: - Use : as the field separator
• $1 == "visitors" - If the first field matches visitors execute the following statement (within { ... })
• size = split($4, users, ","); - Split the 4:th field on , and store the result in the array users. Returns the number of elements in users and puts that in size.
• The for loop - Loop over all the elements in users and print them out.
• exit - Terminate awk - There's no need to continue reading the rest of the file.