Find Default user

Shell Programming and Scripting
1

All,

Working in Kubuntu 14.04

I know I can find the default user line using the cmd from bash :

cat /etc/passwd | grep 1000

What I get is:

user:x:1000:1000:User Name,,,:/home/user:/bin/bash

How do I extract to get:

myuser=$user
myhome=$homdir

All help appreciated!

Thanks!

OMR

2

See if any of these will get you going:

awk -F":" '/1000/ {print $1, $6}' /etc/passwd

or:

awk -F":" '$3=="1000" {print $1, $6}' /etc/passwd

By the way

cat /etc/passwd | grep 1000

is like eating macaronis using a fork first to load into a spoon.
grep can read files by itself

grep 1000 /etc/passwd
3

Expanding on what Aia said to get those values assigned to the desired variables:

read myuser myhome <<< $(awk -F":" '$3=="1000" {print $1, $6}' /etc/passwd)
4

pure bash :

 while IFS=":" read user _ userid _ _ homedir _; do [ $userid -eq 1000 ] && break; done < /etc/passwd

or

IFS=":" read user _ userid _ _ homedir _ <<< $(grep ".*:.*:1000:" /etc/passwd)

This is not false positive proof, though.

5

Rudi, your second solution must have the here-string in quotes.
Just like the following

IFS=":" read myuser x x x x myhome x <<< "`getent passwd 1000`"

Otherwise special characters can spoil it.
And bash seems to have a parsing problem...?

6

Works for my bash . What special chars could spoil it? The here string would supply one single (hopefully) line that would be read into the variables.

7

I get this:

% cat passwd
root:x:0:0:Super-User:/:/bin/sh
user:x:1000:0:Super-User:/:/bin/sh
% IFS=":" read user _ userid _ _ homedir _ <<< $(grep ".*:.*:1000:" passwd); echo $user; echo $userid
user x 1000 0 Super-User / /bin/sh

% IFS=":" read user _ userid _ _ homedir _ <<< "$(grep ".*:.*:1000:" passwd)"; echo $user; echo $userid
user
1000
% echo $BASH_VERSION
3.2.57(1)-release
%