[python-win32] Error message?

[David Kelly]

I got the following error message in stepping through a Python 2.2 script.
Can anyone shed some light for me on what it means?  I am new to Python.

PythonWin 2.2.3 (#42, May 30 2003, 18:12:08) [MSC 32 bit (Intel)] on win32.
Portions Copyright 1994-2001 Mark Hammond (mhammond at skippinet.com.au) - see
'Help/About PythonWin' for further copyright information.
[Dbg]>>> Traceback (most recent call last):
  File
"C:\Python22\lib\site-packages\Pythonwin\pywin\framework\dbgcommands.py",
line 73, in OnGo
    self._DoOrStart("do_set_continue", scriptutils.RS_DEBUGGER_GO)
  File
"C:\Python22\lib\site-packages\Pythonwin\pywin\framework\dbgcommands.py",
line 57, in _DoOrStart
    method()
  File "C:\Python22\lib\site-packages\Pythonwin\pywin\debugger\debugger.py",
line 637, in do_set_continue
    if self.GUIAboutToRun():
  File "C:\Python22\lib\site-packages\Pythonwin\pywin\debugger\debugger.py",
line 788, in GUIAboutToRun
    if not self.StopDebuggerPump():
  File "C:\Python22\lib\site-packages\Pythonwin\pywin\debugger\debugger.py",
line 486, in StopDebuggerPump
    if self.GUIAboutToFinishInteract():
  File "C:\Python22\lib\site-packages\Pythonwin\pywin\debugger\debugger.py",
line 837, in GUIAboutToFinishInteract
    self.oldForeground.EnableWindow(1)
win32ui: The window handle does not specify a valid window
win32ui: Error in Command Message handler for command ID 15022, Code 0

Thanks,

David N. Kelly