I have a program that produces a csv file and right at the end I am using os.startfile(fileName) but then due to the program finishing execution the opening file just closes also, same happens if I add a sleep after also, file loads up then once the sleep ends it closes again?
Any help would be appreciated.
From the documentation for os.startfile:
startfile() returns as soon as the associated application is launched. There is no option to wait for the application to close, and no way to retrieve the application’s exit status.
When using this function, there is no way to make your script wait for the program to complete because you have no way of knowing when it is complete. Because the program is being launched as a subprocess of your python script, the program will exit when the python script exits.
Since you don't say in your question exactly what the desired behavior is, I'm going to guess that you want the python script to block until the program finishes execution (as opposed to detaching the subprocess). There are multiple ways to do this.
The subprocess module allows you to make a subprocess call that will not return until the subprocess completes. The exact call you make to launch the subprocess depends heavily on your specific situation, but this is a starting point:
subprocess.Popen(['start', fileName], shell=True)
You can have your script block until the user tells the python script that the external program has closed. This probably requires the least modification to your code, but I don't think it's a good solution, as it depends on user input.
os.startfile(fileName)
input('Press enter when external program has completed...')
