How to get via SSH the active DISPLAY for a locally logged-in user?

Question

I need to get the active DISPLAY for any logged in user, so if they enter through SSH, they can query it.

Under Ubuntu 12.04, I used a script that, using the consolekit dbus interface, iterated through the active sessions, matching the user UID. A bit complicated, but that would give me the Display that I needed.

This is the code I used:

function obtener_display(){
        _UID=$1
        SESIONES_RAW=$(dbus-send --system --dest=org.freedesktop.ConsoleKit --print-reply /org/freedesktop/ConsoleKit/Manager org.freedesktop.ConsoleKit.Manager.GetSessions)
        SESIONES=$(echo "$SESIONES_RAW" | grep "object path" | sed -r 's/^.*"(.*)".*$/\1/')
        for SESION in $SESIONES ; do
                USUARIOS_RAW=$(dbus-send --system --dest=org.freedesktop.ConsoleKit --print-reply $SESION org.freedesktop.ConsoleKit.Session.GetUnixUser)
                USUARIOS=$(echo "$USUARIOS_RAW" | grep "uint32" | sed -r 's/^.*uint32 (.*)$/\1/')
                for USUARIO in $USUARIOS ; do
                        if [ $_UID -eq $USUARIO ] ; then
                                X11DISPLAY_RAW=$(dbus-send --system --dest=org.freedesktop.ConsoleKit --print-reply $SESION org.freedesktop.ConsoleKit.Session.GetX11Display)
                                X11DISPLAY=$(echo "$X11DISPLAY_RAW" | grep "string" | sed -r 's/^.*string "(.*)"$/\1/')
                                if [ ! -z "$X11DISPLAY" ] ; then
                                        DISPLAY_VALIDO="$X11DISPLAY"
                                fi
                        fi
                done
        done

        if [ ! -z "$DISPLAY_VALIDO" ] ; then
                echo "$DISPLAY_VALIDO"
        else
                echo "FALSE"
        fi
}

And it worked like a charm! Well, kind of. At least it seemed to work properly, never heard anything bad about it. Anyways, it doesn't work under 14.04. The first method I used to get the ConsoleKit sessions returns an empty array. And every method from Manager also return empty arrays.

Is there any way I can fix this? Fix this method should do it, but any other way to get the active DISPLAY for any logged in user should also work.

It does indeed returning an empty array. BTW what does `env` print out at the ssh-console? Doesn't include the active display information? — Khurshid Alam, Jan 07 '15 at 07:20
One other thing you can do is coping display information to a file at boot. For that put `env | grep DISPLAY > $HOME/.active_display` at the end of the `.bashrc` file of the remote machine you want to ssh into. Next time the remote user boots Display information will be available in `.active_display` file. — Khurshid Alam, Jan 07 '15 at 07:26
No, env doesn't give me anything. I ended up with a much simpler solution. I'll post it as an answer. — Jorge Suárez de Lis, Jan 09 '15 at 08:34

score 1 · Answer 1 · edited Feb 26 '17 at 23:17

1

The command who gives me information about current logged in users, and their VTs and displays right away. All I need to do is find the user I want and parse the output.

who | grep $USER | grep -v tty | grep -v pts/ | cut -d' ' -f2

This will output :0, that's what I needed.

edited Feb 26 '17 at 23:17

Zanna

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answered Jan 09 '15 at 08:36

Jorge Suárez de Lis

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your command gives empty result. And `who` doesn't print the word `tty` on 14.04. However this works for me: `who | grep $USER | grep -v pts/ | awk -F' ' '{print $2}'` – Khurshid Alam Jan 09 '15 at 13:04
regarding the above comment, the exact output depends on the terminal emulator if I'm not wrong. (in MATE terminal... `who | sed 's/.*($.*$)$/\1/'` perhaps... Anyway, the `who` command does seem reliably to have the information you want, so +1 – Zanna Feb 26 '17 at 23:19

How to get via SSH the active DISPLAY for a locally logged-in user?

1 Answers1