Why would an "command not recognized" error occur only when a window is populated?
My record sheet app has a menu option for creating a new, blank record sheet. When I open a sheet window, I can open new windows without a problem, using subprocess.Popen() to do it.
However, under Windows (I haven't tested it on other OSes yet), if I open a new window then use the "open file" dialog to populate the fields with data from a file, I'm no longer able to create new windows. Once it's populated, Windows gives me the
'foo.py' is not recognized as an internal or external command, operable program or batch file.
I don't understand what would cause Windows to suddenly not recognize the Popen() call. I don't have any code that would affect it in any way that I'm aware of.
Asked by: Elian334 | Posted: 28-01-2022
Answer 1
From the error message, it looks like you need to pass the full path of "foo.py" to your Popen call. Normally just having "foo.py" will search in your current working directory, but this can be a bit unpredictable on Windows, I have found. Yours seems to be jumping around with the open file dialog.
Secondly, just for good measure, it would seem like you would need to pass foo.py as an argument to python.exe executable, rather than executing foo.py itself. Again, I would specify this by path.
So to be safe, something like:
subprocess.Popen([r'C:\Python2.5\python.exe', r'C:\path\to\foo.py'])
Answer 2
The suggested answer seems to have fixed the problem. I also realized that I needed to use os.name to determine which OS is being used, then I can use the correct path format for loading the external Python file.
Answered by: Lydia924 | Posted: 01-03-2022Similar questions
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