Echo only valid user

Bash newbie question: I’d like to have this script echo out only users from a given list that are valid user IDs, and not echo the invalid ones.

Here’s what I have so far:

#!/bin/bash

while IFS= read -r line
    do
        id "$line"

        if [ $? -eq 1 ] ; then

        echo $line

        else

        echo "$line is not valid user ID" >&2
fi
    done < "$1"

Instead, it is echoing out the results of ‘id’ as well as the text “is not valid user ID”

uid=17931(wwirls) gid=100000(all_usr) groups=100000(all_usr),12(everyone),62(netaccounts),701(1)
wwirls is not valid CruzID
id: smak: no such user
smak

Ideally, it’d echo results like:

wwirls
otheruserid
admin

Your help is appreciated!

Answer

id "$line" &> /dev/null

is probably the smallest change that will do what you want. That says to send the output of id (including any error messages) to the bit bucket (/dev/null). However, the sense of your test is backwards – $? is 0 on success, and 1 on failure.

Now, as it happens, if automatically looks at $?. So you can shorten this to:

if id "$line" &> /dev/null    # Find out if the user exists, without printing anything
then
    echo "$line"              # The user exists (`id` succeeded)
else
    echo "$line is not valid user ID" >&2     # No such user
fi

Edit 2 If you don’t need the not valid output, you can replace the whole if..fi block with

id "$line" &> /dev/null && echo "$line"

If the command before && succeeds, bash runs the command after &&.