On Windows XP, I'm able to launch the application associated with file1.xxx.
os.startfile("file1.xxx")
I want to start more instances of the application, so that I can process
several files at the same time.
calling os.startfile("file2.xxx") causes the instance of the application to
also open file2.xxx.
The problem with this is- I have to wait until file1 has finished
processing, before I can start file2.
If, after opening the first instance for file1, I can use the Start Menu,
Program Files, open another instance of the application, I can then start
processing file2 at the same time as file1, but using a separate instance.
Is there any other way to invoke startfile() or other function- to force a
new instance of the application to open?
How would I close the application from Python, that was started with
os.startfile()?
Once the application is open, I just want to use File Save As Text, to get
the data into a text file.
Python's app-automation capabilities are kinda weak and not well documented,
so I may resort to AutoIt3.
Thanks
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