Script to validate home dir

Question

I want to write a simple Bash script which displays the path to the home directory of a user when an actual user name is used as an argument and displays user not found or something when a user name that doesn't exist on this system is used as an argument.

Would this require a function? Can this be done by using basic GNU compatible commands in a script file?

Answer 1

read -p "Enter a username: " username
if getent passwd "$username" > /dev/null
then
  printf "Their home directory is: %s\n" "$(getent passwd "$username" | cut -d: -f6)"
else
  printf "User not found!\n" >&2
fi

Answer 2

#!/bin/sh

username=$1

if ! getent passwd "$username" >/dev/null 2>&1; then
    printf 'User %s does not exist\n' "$username"
    exit 1
fi

homedir=$( getent passwd "$username" | cut -d: -f6 )

if [ -n "$homedir" ]; then
    if [ -d "$homedir" ]; then
        printf 'User %s has a valid (existing) home directory: %s\n' "$username" "$homedir"
    else
        printf 'User %s lacks a valid (existing) home directory: %s\n' "$username" "$homedir"
    fi
else
    printf 'User %s has no home directory\n' "$username"
fi

The script takes the username from the command line:

$ ./script.sh kk
User kk has a valid (existing) home directory: /home/kk

$ ./script.sh nobody
User nobody lacks a valid (existing) home directory: /nonexistent

$ ./script.sh aoae
User aoae does not exist

Answer 3

getent passwd <USERNAME> is a good way to read the user database. It prints the entry from the user database for the given user, if the user exists, or exits non-zero if the user doesn't exist.

So we can look at the output if it exited zero, and print an error message if it exited non-zero.

The home directory is the 6th field of the passwd entry. It can be extracted using standard text processing tools, e.g. awk or cut.

if passwd="$(getent passwd "$user")"; then
    awk -F: '{print $6}' <<<"$passwd"
else
    echo "No such user $user" >&2
fi