如何从/ etc / passwd格式化用户帐户的输出

Question

I want to format output of a user account from /etc/passwd to display only the name, role, and directory path, all separated by commas. I know this should be easy, but for the life of me I cannot figure out how to display between certain colons. (Note this should work with any username, not just the ex)

Ex of grep joe /etc/passwd :

joe:x:1001:1001:System Admin:/home/joe:/bin/bash

Desired Output :

joe, System Admin, /home/joe

Thank you!

Answer 1

awk 'BEGIN{ FS=":"; OFS=", " } $1=="joe" { print $1,$5,$6; exit }' /etc/passwd 

（但是下次您应该付出更多的努力-您的问题非常令人讨厌：）

Answer 2

使用cut将逗号用作--output-delimiter ：

cut -d: -f1,5,6 --output-delimiter=, /etc/passwd

Answer 3

Try:

grep joe /etc/passwd | cut -d: -f1,5,6

If you really need "," as a delimiter:

grep "^joe:" /etc/passwd | cut -d: -f1,5,6 | tr : ,

The "^" ensures only matches to "joe" at the beginning of the line are intercepted (Thanks PSkocik for the reminder). grep by defaults accepts a regex .

Translation in bash, but two lines:

x=`grep joe /etc/passwd | cut -d: -f1,5,6`
echo ${x/:/,}

Answer 4

If you want to do it purely in shell (POSIX sh is enough, Bash not needed), then read is your friend:

while IFS=: read user pw uid gid gecos home shell ; do 
    if [ "$user" = "joe" ] ; then 
        echo "$user, $gecos, $home" 
    fi 
done < /etc/passwd

Answer 5

Adding to the variations, in bash itself, you can pipe the output of grep directly to a brace enclosed read and separate the fields using IFS as required. Using short names for the fields, you could do:

grep '^joe:' /etc/passwd | 
{ IFS=: read -r u x uid gid n h s; echo "$u, $n, $h"; }

Which would give the output you seek (if /etc/passwd contained the entry)