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 ([EMAIL PROTECTED]) - 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
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