Finding duplicate users on CentOS

Question

I'm using the following code to find the duplicate username. However, it gives an error.

#!/bin/bash
cat /etc/passwd | cut -f1 -d":" | /bin/sort -n | /usr/bin/uniq -c |\
    while read x; do [ -z "${x}" ] && break set - $x
        if [ $1 -gt 1 ]; then
            uids=`/bin/gawk -F: '($1 == n) { print $3 }' n=$2 \
            /etc/passwd | xargs`
            echo "Duplicate User Name ($2): ${uids}"
        fi
    done

I'm facing a syntax error near the token 'done' and numeric error. How can I fix this error?

Answer 1

$ cut -d: -f1 /etc/passwd | sort | uniq -d

This will extract the first field (the usernames) of the :-delimited /etc/passwd-file, sort the result and report any duplicates.

To also get the UID and the rest of the duplicated passwd entries:

cut -d: -f1 /etc/passwd | sort | uniq -d |
while read -r username; do
  grep "^$username:" /etc/passwd
done

To only get the duplicate usernames and their UID:

cut -d: -f1 /etc/passwd | sort | uniq -d |
while read -r username; do
  awk -F: -vu="$username" '$1 == u { print $1, $3 }' /etc/passwd
done

A short note on your script. The syntax looks mostly ok, but you need ; after break and there is a space after both \ (this may be a cut-and-paste error (now removed by an edit)). Also, I'd avoid giving full paths to standard utilities if there is no good reason for it, and the awk program does not require GNU awk so just awk will do.

Answer 2

1. You need to have a do somewhere between the while and the done — typically, right after the read, or after the check that you got data.
2. set - $x should be on a line by itself, or at least separated from the break with a semicolon (;).  (This is probably a good place to put the do.)

Suggestions:

3. Rather than doing set - $x, consider changing read x to read count name.
4. Just for clarity, you might want to change `…` to $(…) — see this, this, and this.

Answer 3

I would do something like that with awk (one-liner):

awk -F: '{if ($1 in users) print "Duplicate Username: "$1 ; else users[$1]}' /etc/passwd

find username in users array variable, if duplicate print msg else add the user to the array

Answer 4

Inspired by ebal’s answer, this single awk command replicates the intended functionality of the OP’s script:

awk -F: '{ uids[$1] = uids[$1] " " $3; }
         END { for (u in uids) { if (uids[u] ~ /. /)
                         print "Duplicate User Name (" u "):" uids[u]; } }'

As in ebal’s answer, it creates an associative array, with an entry for each unique user name.  It sets the value of each entry to the concatenation of a space and the UID for each line for the user.  Then, after it’s read the entire file, it loops through the unique user names and checks whether it occurs more than once — recognized by the fact that the UID list has a space as other than the first character — and reports the desired information.

This will behave oddly if the UIDs in the file contain spaces.