Opening a simple file using OS in python

Asked Mar 14 '18 at 02:17

Active Mar 14 '18 at 02:21

Viewed 52 times

Python code

import os
def open_ss():
    os.startfile('C:\\WINDOWS\\system32\\SnippingTool.exe')

Error message

File "<stdin>", line 1, in <module>
  File "C:\Users\Jong\Desktop\Projects\MiniScript\main.py", line 5, in open_ss
    os.startfile('C:\\WINDOWS\\system32\\SnippingTool.exe')
FileNotFoundError: [WinError 2] The system cannot find the file specified: 'C:\\WINDOWS\\system32\\SnippingTool.exe'

Why is the file specified showing up as "C:\\WINDWOS\\system\\SnippingTool.exe" instead of just "C:\WINDWOS\system\SnippingTool.exe"

edited Mar 14 '18 at 02:21

ndmeiri

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asked Mar 14 '18 at 02:17

Jong Woong To

use subprocess.call – Arkadiusz Tymieniecki Mar 14 '18 at 02:23
i get the same error – Jong Woong To Mar 14 '18 at 02:25
In Python the backslash character is an escape character. So when using it in a file path, it too has to be escaped in order to find a file, hence the double backslashes. – Michael Swartz Mar 14 '18 at 02:26
So how can I fix the url 'C:\\WINDOWS\\system32\\SnippingTool.exe' inorder for it to work? – Jong Woong To Mar 14 '18 at 02:28
Make it a raw string: `os.startfile(r'C:\WINDOWS\system32\SnippingTool.exe')` – Michael Swartz Mar 14 '18 at 02:54
Its giving me the same error still – Jong Woong To Mar 14 '18 at 04:47
It seems that this particular problem only occurs when trying to open SnippingTool. It works fine when I try to open mspaint or notepad. – Jong Woong To Mar 15 '18 at 00:40

Opening a simple file using OS in python

0 Answers0