Shell Programming and Scripting

i was trying to work on program to look for users never log on sever.. using awk
with awk is working
last| awk '{print $1}' |sort -u > /tmp/users1$$
cat /etc/passwd | awk -F: '{print $1}' |sort -u > /tmp/users2$$
comm -13 /tmp/users[12]$$
rm -f /tmp/users[12]$$

with cut it is not working

last| cut -c1-10 |sort -u > /tmp/users1$$
cat /etc/passwd |cut -d':' -f1 | sort -u > /tmp/users2$$
comm -13 /tmp/users[12]$$
rm -f /tmp/users[12]$$

i have another idea by using for loop and count number users repeated. if anyone has better idea to solve this problem please help.

It isn't working because with cut, you are cutting the first 10 columns. This causes spaces to be padded to the actual usernames in the /tmp/users1$$ file.

Use this instead:

last| cut -d" " -f1 |sort -u > /tmp/users1$$

it worked ..thanks
but is there any other way by not using comm and diff to compare 2 files ..

I love one-line commands.

Code:

(last | awk '{print$1}'; awk -F: '{print$1}' /etc/passwd) | sort | uniq -u

The command "uniq -u" show only unrepeated lines, which would be "users in passwd but NOT in wtmp".