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

Question

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.

Answer 1

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'])

Answer 2

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.

Answer 3

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

Answer 4

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)

Answer 5

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'])

Answer 6

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

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

Answer 7

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")

Answer 8

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'])

Answer 9

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

Answer 10

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)