I have a Python script that needs to execute an external program, but for some reason fails.
If I have the following script:
import os; os.system("C:\\Temp\\a b c\\Notepad.exe"); raw_input();
Then it fails with the following error:
'C:\Temp\a' is not recognized as an internal or external command, operable program or batch file.
If I escape the program with quotes:
import os; os.system('"C:\\Temp\\a b c\\Notepad.exe"'); raw_input();
Then it works. However, if I add a parameter, it stops working again:
import os; os.system('"C:\\Temp\\a b c\\Notepad.exe" "C:\\test.txt"'); raw_input();
What is the right way to execute a program and wait for it to complete? I do not need to read output from it, as it is a visual program that does a job and then just exits, but I need to wait for it to complete.
Also note, moving the program to a non-spaced path is not an option either.
This does not work either:
import os; os.system("'C:\\Temp\\a b c\\Notepad.exe'"); raw_input();
Note the swapped single/double quotes.
With or without a parameter to Notepad here, it fails with the error message
The filename, directory name, or volume label syntax is incorrect.
subprocess.call will avoid problems with having to deal with quoting conventions of various shells. It accepts a list, rather than a string, so arguments are more easily delimited. i.e.
import subprocess subprocess.call(['C:\\Temp\\a b c\\Notepad.exe', 'C:\\test.txt'])
Here's a different way of doing it.
If you're using Windows the following acts like double-clicking the file in Explorer, or giving the file name as an argument to the DOS "start" command: the file is opened with whatever application (if any) its extension is associated with.
filepath = 'textfile.txt' import os os.startfile(filepath)
import os os.startfile('textfile.txt')
This will open textfile.txt with Notepad if Notepad is associated with .txt files.