How do I execute a program from Python? os.system fails due to spaces in path

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)

Example:

import os
os.startfile('textfile.txt')

This will open textfile.txt with Notepad if Notepad is associated with .txt files.

The outermost quotes are consumed by Python itself, and the Windows shell doesn't see it. As mentioned above, Windows only understands double-quotes. Python will convert forward-slashed to backslashes on Windows, so you can use

os.system('"C://Temp/a b c/Notepad.exe"')

The ' is consumed by Python, which then passes "C://Temp/a b c/Notepad.exe" (as a Windows path, no double-backslashes needed) to CMD.EXE

At least in Windows 7 and Python 3.1, os.system in Windows wants the command line double-quoted if there are spaces in path to the command. For example:

  TheCommand = '\"\"C:\\Temp\\a b c\\Notepad.exe\"\"'
  os.system(TheCommand)

A real-world example that was stumping me was cloning a drive in VirtualBox. The subprocess.call solution above didn't work because of some access rights issue, but when I double-quoted the command, os.system became happy:

  TheCommand = '\"\"C:\\Program Files\\Sun\\VirtualBox\\VBoxManage.exe\" ' \
                 + ' clonehd \"' + OrigFile + '\" \"' + NewFile + '\"\"'
  os.system(TheCommand)

For python >= 3.5 subprocess.run should be used in place of subprocess.call

https://docs.python.org/3/library/subprocess.html#older-high-level-api

import subprocess
subprocess.run(['notepad.exe', 'test.txt'])

import win32api # if active state python is installed or install pywin32 package seperately

try: win32api.WinExec('NOTEPAD.exe') # Works seamlessly
except: pass

I suspect it's the same problem as when you use shortcuts in Windows... Try this:

import os;
os.system("\"C:\\Temp\\a b c\\Notepad.exe\" C:\\test.txt");

Suppose we want to run your Django web server (in Linux) that there is space between your path (path='/home/<you>/<first-path-section> <second-path-section>'), so do the following:

import subprocess

args = ['{}/manage.py'.format('/home/<you>/<first-path-section> <second-path-section>'), 'runserver']
res = subprocess.Popen(args, stdout=subprocess.PIPE)
output, error_ = res.communicate()

if not error_:
    print(output)
else:
    print(error_)

---

[Note]:

• Do not forget accessing permission: chmod 755 -R <'yor path'>
• manage.py is exceutable: chmod +x manage.py

For Python 3.7, use subprocess.call. Use raw string to simplify the Windows paths:

import subprocess
subprocess.call([r'C:\Temp\Example\Notepad.exe', 'C:\test.txt'])

No need for sub-process, It can be simply achieved by

GitPath="C:\\Program Files\\Git\\git-bash.exe"# Application File Path in mycase its GITBASH
os.startfile(GitPath)