Sed $DISPLAY of currently logged in user

Question

I want to find out the used DISPLAY of the currently logged in user. For that I wanted to use sed. First, the output of who:

[orschiro@thinkpad ~]$ who
orschiro tty1         2013-08-05 23:15
orschiro pts/0        2013-08-05 23:17 (:0)
orschiro pts/1        2013-08-05 23:22 (:0)
orschiro pts/2        2013-08-05 23:22 (:0)

That is I want to retrieve :0 for the logged in user orschiro.

I am using the following expression but it does not retrieve the expected result. Instead the output is empty:

[orschiro@thinkpad ~]$ who | sed -e "/orschiro/! d;/pts/d;s/^.*[^0-9]\\(:[0-9.]\\+\\).*$/\\1/p;d" | head -n1
[orschiro@thinkpad ~]$ 

What is wrong with my expression?

Answer 1

Command /pts/d; delete all lines, that contain string pts. Try this:

who | sed -e '/orschiro/! d; /pts/! d; s/^.*\(:[0-9.]\+\).*$/\1/p;d' | head -n1

Answer 2

Maybe this?

who | awk -F '[()]' '/orschiro/{print $(NF-1)}' | grep -v orschiro | head -n1

or

who | awk -F '[()]' '/orschiro/{print $(NF-1)}' | grep -v orschiro | uniq

Answer 3

Try this one:
who | awk '/orschiro/{print $5}' | sed -e 's/[()]//g'