What is the group id of this group name?

command-line groups

126,559

Solution 1

A hack for the needed: (still maybe there is much better answer)

awk -F\: '{print "Group " $1 " with GID=" $3}' /etc/group | grep "group-name"

Simpler version(Thanks to @A.B):

awk -F\: '/sudo/ {print "Group " $1 " with GID=" $3}' /etc/group

Example:

$ awk -F\: '{print "Group " $1 " with GID=" $3}' /etc/group | grep sudo 
Group sudo with GID=27

Solution 2

Use the getent command for processing groups and user information, instead of manually reading /etc/passwd, /etc/groups etc. The system itself uses /etc/nsswitch.conf to decide where it gets its information from, and settings in the files may be overridden by other sources. getent obeys this configuration. getent prints data, no matter the source, in the same format as the files, so you can then parse the output the same way you would parse /etc/passwd:

getent group sudo | awk -F: '{printf "Group %s with GID=%d\n", $1, $3}'

Note that, for a username, this much more easier. Use id:

$ id -u lightdm
105

Solution 3

This can be simply done with cut:

$ cut -d: -f3 < <(getent group sudo)
27

getent group sudo will get the line regarding sudo group from /etc/group file :

$ getent group sudo
sudo:x:27:foobar

Then we can just take the third field delimited by :.

If you want the output string accordingly, use command substitution within echo:

$ echo "Group sudo with GID="$(cut -d: -f3 < <(getent group sudo))""
Group sudo with GID=27

126,559

Maythux

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Updated on September 18, 2022

Comments

Maythux over 1 year
How to get the group ID GID giving the group name.

The output would be for example:
```
Group adm with GID=4
```
A.B. almost 9 years

Try this: awk -F\: '/sudo/ {print "Group " $1 " with GID=" $3}' /etc/group
A.B. almost 9 years

Ok, awk is shorter than perl =)
muru almost 9 years

Pretty sure perl can do the splitting for you, I think the -F or -l option does that.
A.B. almost 9 years

% echo "Group cdrom with GID="$(cut -d: -f3 < <(getent group sudo))"" Group cdrom with GID=27, please a little more generic =)
kos almost 9 years

Why the process substitution? What's wrong with GID="$(getent group cdrom | cut -d: -f3)"?
kos almost 9 years

See UUOC
heemayl almost 9 years

@kos I don't like running things in subshells unless is a must.....
heemayl almost 9 years

@A.B. Edited...
kos almost 9 years

But it's POSIX...:'( No, just joking, I genuinely didn't know the difference. +1
Zelphir Kaltstahl about 2 years

Note, that this is Bash specific. Same result can be achieved with: getent group root | cut --delimiter ':' --fields 3 (or use short args). Seems like this case does not really require process substitution.