How do I list logged-in users without duplicates?

Solution 1:

We can pipe the output of who to awk to print only the first cell of each record (row) and then pipe it to the command sort, that will sort the values alphabetically and will output only the unique -u entries:

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

Or we can use only awk in this way:

who | awk '!seen[$1]++ {print $1}'

A POSIX compliant solution, provided by @dessert - where cut will use the spaces as delimiter -d' ' and will print only the first field of each record -f1:

who | cut -d' ' -f1 | sort -u

Thanks to @DavidFoerster here is a lot shorter syntax that doesn't lose the information of all the other columns:

who | sort -u -k 1,1

For the same purposes we could use the command w with the option -h (ignore headers), for example:

w -h | awk '!seen[$1]++ {print $1}'

We could use also the command users combined with the command rs (reshape data) with the transpose option -T and then again sort -u:

users | rs -T | sort -u

We could use and who -q with transposition in the following way - where the command head -1 will crop only the first line of the output of the previous command:

who -q | head -1 | rs -T | sort -u

See also:

• How do I find who is logged-in as root?

• How do I get the list of the active login sessions?