Eliminating duplicates from the who command

Shell Programming and Scripting
1

I am trying to show how many users are logged into one of my systems. Using who -q it gives me a user count, but some users are logged in multiple times.

Is there any easy parameter that I can use to ignore/eliminate these duplicates??

Thanks

2

Can you try with who|uniq , see whether you are get rid of dups.

Cheers,
gehlnar

3

No... no joy :-(,

who|uniq outputted exactly the same as the who command without the pipe.

4 
who | awk '{ print $1 }' | sort -u
5

Hi,

This one also can help.

who | awk '{print $1}' | uniq

6 
who | cut -f1 -d' ' | sort | uniq
7

Thanks, that has listed all the unique users, how would I then calculate the amount of users, like who -q shows #users... any thoughts?

8 
who -q | wc -w
9

This should give the number of unique users:

who|awk '!a[$1]++{print $1;i++}END{print "users = "i}'

Regards

10

Just pipe any of the previous solutions to wc -l, for example:

who | awk '{ print $1 }' | sort -u | wc -l