I have an app running remotely on a beaglebone, written in wxPython. I want to run the app inside of a virtual frame buffer, using Xvfb, and access it remotely using a VNC.

It may also be pertinent for you to know that I am launching VNC from inside of the application, like so:

 display = Display(":99") root = display.screen().root win_ids = [w.id for w in root.query_tree()._data['children'] if w.get_wm_name() and 'myappname' in w.get_wm_name().lower()] if (win_ids[0]): print("Found my win id. Starting x11vnc") system("x11vnc -quiet -sid %s -forever &" % win_ids[0]) else: print("Can't find my Window ID!") 

When I launch the app with:

 sudo xvfb-run ./myApp -s :99 -screen 0 1920x1080x16 -ac 

I see the following:

However, when I launch Xvfb separately:

 Xvfb :99 -screen 0 800x600x16 -ac & 

and then launch the app:

 DISPLAY=:99 ./myApp 

I see this (what it should be):

Any insights?

I have an app running remotely on a beaglebone, written in wxPython. I want to run the app inside of a virtual frame buffer, using Xvfb, and access it remotely using a VNC.

It may also be pertinent for you to know that I am launching VNC from inside of the application, like so:

 display = Display(":99") root = display.screen().root win_ids = [w.id for w in root.query_tree()._data['children'] if w.get_wm_name() and 'myappname' in w.get_wm_name().lower()] if (win_ids[0]): print("Found my win id. Starting x11vnc") system("x11vnc -quiet -sid %s -forever &" % win_ids[0]) else: print("Can't find my Window ID!") 

EDIT: I am using Debian Jessie with kernel version 3.8.13, and xenomai. I can't find a version number for xvfb-run or Xvfb, but I can tell you that my xorg version is 1.16.4

When I launch the app with:

 sudo xvfb-run ./myApp -s :99 -screen 0 1920x1080x16 -ac 

I see the following:

However, when I launch Xvfb separately:

 Xvfb :99 -screen 0 800x600x16 -ac & 

and then launch the app:

 DISPLAY=:99 ./myApp 

I see this (what it should be):

Any insights?