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: 
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

```
Maybe this?

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

```

or

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

```