How to count the number of active windows open apart from that of the window manager?

Question

I would like to count the number of active windows open apart from that of the window manager I am using. I am aware that wmctrl can do this, however, when I do the following:

wmctrl -l

on my Ubuntu 14.04 64-bit desktop using unity as the window manager, the output I get is:

0x03c00002  0 stepup XdndCollectionWindowImp
0x03c00005  0 stepup unity-launcher
0x03c00008  0 stepup unity-panel
0x03c0000b  0 stepup unity-dash
0x03c0000c  0 stepup Hud
0x02c0000a  0 stepup Desktop
0x02e000b3  0 stepup How to count the number of active windows open irrespective of window manager? - Unix & Linux Stack Exchange - Mozilla Firefox
0x0340000c  0 stepup Terminal

However, I would want to ignore the windows that unity/gnome/KDE, etc. uses and count just the windows the user has opened on his own. How do I this?

Answer 1

There is no formal definition of window belonging to the window manager or “opened on his own”. Technically, the window manager role doesn't call for any window: it's other parts of the desktop environment (typically called widgets) that have windows. There is no attribute on a window or on an application that says “this is part of the desktop environment”.

You can run wmctrl -lx and identify the windows you don't care about by their class. I haven't checked what classes the various common desktop environments use, fill in the list as desired.

wmctrl -lx | awk '$3 !~ /^(Unity-.*|Kwin|Boringstuff)$/'

Alternatively, maybe detecting non-sticky windows will fit your bill. Sticky windows are the ones that are displayed on all workspaces. I don't know if this works with all window managers.

for w in $(wmctrl -l | awk '{print $1}'); do
  if ! xprop -id $w _NET_WM_STATE | grep -q _NET_WM_STATE_STICKY; then
    wmctrl -l -i $w
  fi
done