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I want to list the username, id and group on /etc/passwd using the following format: username uid gid
I have used the following:
cut -d: -f1,3,4 /etc/passwd
But it returns username:uid:gid. How can I format the command to remove the : or list without it, like this:
root 0 0
daemon 1 1
bin 2 2
...
Depending on what output you want
The simplest way is to translate the delimiter to what you want with tr, sed or awk... For example
cut -d: -f1,3,4 /etc/passwd | tr ':' '\t'
cut -d: -f1,3,4 /etc/passwd | sed 's/:/ --- /g'
awk -F: '{ print $1, $3, $4}' /etc/passwd
If you want to format as table then use column
cut -d: -f1,3,4 /etc/passwd | column -t -s ':'
My version of cut (cut (GNU coreutils) 8.28) has a --output-delimiter argument:
cut -d: -f 1,3,4 --output-delimiter " " /etc/passwd
result:
root 0 0
daemon 1 1
bin 2 2
sys 3 3
/etc/passwd(or/etc/group) directly. Usegetent passwd(orgetent group) instead. /etc/passwd is only one of many possible sources of data for user account info. ditto for /etc/group. See the man pages fornss,getent, andnsswitch.conf. e.g. using @phuclv's awk example:getent passwd | awk -F: '{print $1, $3, $4}'getentinstead of a comment. That way people can upvote it, and it can potentially be selected as the correct answer to achieve what the OP wants.getentinstead of reading /etc/passwd directly.