I'm trying to open any program file via python

Question

I'm trying to open a file via my python project, but it doesn't work.. Is there any reason why?

import os
from tkinter import *
from tkinter import filedialog
from PIL import ImageTk,Image

def openfile():

   window.filename = filedialog.askopenfilename(initialdir="Open file", filetypes=(("exe files", "*.exe"),("all files", "*.*"))
    file_opener = open(window.filename)
    file_opener.read()

Button = Button(window, text="Open", command=openfile)
Button.pack()

Answer 1

To run a file, use os.system([file path]) . That runs the command [file path] which runs the file.

For example, if you have a file in your desktop name test.py , typing os.system("C:\Users\[username]\desktop\test.py") runs the file test.py .

There is a code which runs a file chosen by user:

from tkinter.filedialog import *
import os

path = askopenfilename()

os.system(path)

But if the path contains spaces, add the following code just before os.system(path) :

for character in path:
    if character == ' ':    # space detected?
        path = ('\"'        # add " at the beginning
               + path       # add the path
               + '\"')      # add " at the end
        break

print(path)                 # check if the path is corrected