How can one get executable file-path for a window which is launched as FlatPak application?

Question

I know that

xprop | awk '/PID/ {print $3}' | xargs ps

and

xprop | awk '/COMMAND/ {print $4}' | xargs which

work great with GUI applications which are deb-packaged, AppImage'd or Snap'd.

But this method does not give executable file-name for application which is packaged as FlatPak.
See example below:

$ flatpak run org.libretro.RetroArch

$ xprop 
_NET_WM_VISIBLE_NAME(UTF8_STRING) = "RetroArch   (as superuser)"
WM_NAME(STRING) = "RetroArch  "
_NET_WM_ICON_GEOMETRY(CARDINAL) = 446, 744, 207, 24
WM_PROTOCOLS(ATOM): protocols  WM_DELETE_WINDOW
XKLAVIER_STATE(INTEGER) = 0, -180379392
_NET_WM_STATE(ATOM) = 
WM_STATE(WM_STATE):
      window state: Normal
      icon window: 0x0
_NET_FRAME_EXTENTS(CARDINAL) = 3, 3, 23, 3
_NET_WM_DESKTOP(CARDINAL) = 0
_NET_WM_ALLOWED_ACTIONS(ATOM) = _NET_WM_ACTION_MOVE, _NET_WM_ACTION_RESIZE, _NET_WM_ACTION_FULLSCREEN, _NET_WM_ACTION_MINIMIZE, _NET_WM_ACTION_SHADE, _NET_WM_ACTION_MAXIMIZE_HORZ, _NET_WM_ACTION_MAXIMIZE_VERT, _NET_WM_ACTION_CHANGE_DESKTOP, _NET_WM_ACTION_CLOSE, _NET_WM_ACTION_ABOVE, _NET_WM_ACTION_BELOW
WM_CLIENT_MACHINE(STRING) = "norbert-UX32A"
_NET_WM_PID(CARDINAL) = 2
WM_CLASS(STRING) = "retroarch", "retroarch"
_NET_WM_ICON(CARDINAL) =  Icon (16 x 16):
                  
    ▓▓▓▓▓▓▓▓▓▓▓   
   ▓▓▓▓▓▓▓▓▓▓▓▓▓  
   ▓▓▓▓▓▓▓▓▓▓▓▓▓  
   ▓▓▓ ▓▓▓▓▓ ▓▓▓  
   ▓ ▓▓ ▓▓▓ ▓▓ ▓  
   ▓ ▓       ▓ ▓  
   ▓   ▓   ▓   ▓  
   ▓           ▓  
   ▓▓         ▓▓  
   ▓▓▓ ▓▓▓▓▓ ▓▓▓  
   ▓▓ ▓▓▓▓▓▓▓ ▓▓  
   ▓▓▓▓▓▓▓▓▓▓▓▓▓  
   ▓▓▓▓▓▓▓▓▓▓▓▓▓  
    ▓▓▓▓▓▓▓▓▓▓▓

So I do not see PID- or COMMAND-related lines in xprop output. But of course it is launched:

$ flatpak ps
Instance  PID   Application            Runtime
752290322 23857 org.libretro.RetroArch org.kde.Platform

How can I get PID of window created by FlatPak application while using xprop, wmctrl or something similar?

When using `wmctrl -lp`, my Flatpak apps either show a PID of 0 or of another PID completely unrelated to the app; `flatpak ps`, of course, shows the proper PID. I've looked through **/run/user/$your_UID/.flatpak** and couldn't found anything you might be able to use. — ajgringo619, Dec 23 '19 at 23:38
I also tried `xdotool getwindowpid $window_ID`, and got: `window $window_ID has no pid associated with it.` You might be out of luck. — ajgringo619, Dec 23 '19 at 23:43

score 2 · Accepted Answer · answered Dec 24 '19 at 17:36

In most cases, flatpak permissions (sandboxing) do not allow for this AFAICT. Just how much of the system is exposed is configurable to some degree so which flatpak you are talking about might be germaine to the question. There's a great deal of information on this and other flatpak related topics here.

One notable section is http://docs.flatpak.org/en/latest/sandbox-permissions.html

One alternative approach to locating the executable file path of a flatpak might be checking the configuration files: Excerpt below.

flatpak can operate in system-wide or per-user mode. The system-wide data is located in $prefix/var/lib/flatpak/, and the per-user data is in $HOME/.local/share/flatpak/.

In addition to the default installation locations, more system-wide installations can be defined via configuration files /etc/flatpak/installations.d/, which must have the .conf extension and typically include the installation path in the form Path=/location/of/installation. See the installation documentation for more information.

How can one get executable file-path for a window which is launched as FlatPak application?

1 Answers1