Trying to check whether a list contains an item in bash/ unix without validating substrings but including the first and last items

Question

This is my first time posting here so any feedback is very welcome should I be doing something wrong! I'm very new to shell scripting and am trying to write a function that checks whether an item is in a list. In this case I'm checking whether a given argument correspond to an existing user on the system. I really want to avoid false positives by making sure the item is surrounded by spaces but that obviously excludes the first and last item in the list. I figured maybe I could add spaces to the start and end of the list but I can't figure out how to do this. Any tips or pointers are very welcome! My code so far comes from this post. It looks like this:

-->This is the list of existing users

users=$(getent passwd {1000..60000} | awk -F ':' '{print $1}')

-->Function to check whether the given command line argument is in existing users

contains(){
   [[ $1 =~ (^|[[:space:]])"$2"($|[[:space:]]) ]] && return 0 || return 1
}

Where $1 is the list of users obtained with getent from passwd file (see above) and $2 is the given (command line) argument I'm trying to validate.

Just as information: for testing purposes I've been trying out the contains function with a list (1 2 3 4). It works if I check this against 2 and 3, but not for 1 and 4.

Answer 1

Try the custom function below. It accepts a list where each item is on a separate line. See usage.

contains(){ printf "%s\n" "$1"|grep '^'"$2"'$' > /dev/null; }

The return code is that of the grep command:

• 0 - if the item was found in the list (full line match, case sensitive)
• 1 - if the item was not found in the list
• 2 or higher - if an error occurred (the error message will be printed)

Modifications that you might need to do on the above function to make it work as you want:

• don't print anything, not even a possible error message: &> /dev/null
• print everything (the item if present in the list, a possible error message): remove > /dev/null
• print only if the item is present in the list, don't print a possible error message: 2> /dev/null
• force the function to return only 0 or 1: && return 0 || return 1;

Usage:

list="$(getent passwd {1000..60000}|cut -f1 -d:)"
item="USER_NAME_HERE"
#or
list="$(printf "%s\n" {1..4})"
item="1"
#then
contains "$list" "$item"
#to see the return code
echo $?

Answer 2

users="$(getent passwd | awk -F: '{print $1}')"

contains(){
   grep -qFx "$2" <<< "$1"
}

Test this with something like:

if contains "$users" "jim"; then
  echo jim is a user
else
  echo jim is not a user
fi