Open File with Python

Question

I am writing a tkinter program that is kind of a program that is like a portfolio and opens up other programs also writen in python. So for example i have FILE_1 and FILE_2 and i want to write a program that onced clicked on a certain button opens either FILE_1 or FILE_2. i dont need help with the look like with buttons just how to wirte a function that opens a program

This is the code i used:

from Tkinter import *
import subprocess

master = Tk()

def z():
    p=subprocess.Popen('test1.py')
    p.communicate()


b = Button(master, text="OK", command=z)
b.pack()



mainloop()

unutbu · Accepted Answer · 2010-07-03 20:33:15Z

3

Hook the button up a callback which calls subprocess.Popen:

import subprocess
p=subprocess.Popen('FILE_1.py')
p.communicate()

This will try to run FILE_1.py as a separate process. p.communicate() will cause your main program to wait until FILE_1.py exits.

answered Jul 3, 2010 at 20:33

unutbu

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11 Comments

DonJuma Over a year ago

thanks but what if i want it to have two buttons wouldn't they both open the same file or is there a way to distinguish between files?

unutbu Over a year ago

@Matthew: Each button can be connected to its own callback function. So pressing each button would invoke a different function which could run a different program.

DonJuma Over a year ago

and does this way open it in another window

DonJuma Over a year ago

what i mean is if i call subprocess.Popen it will open FILE_1 and lets say i had another function called G=subprocess.Popen('FILE_2.py') woould i call subprocess.G ?

unutbu Over a year ago

With something like: b = Button(master, text="Run FILE_1", command=file1_callback), clicking Button b would cause file1_callback() to be run. See effbot.org/tkinterbook/button.htm

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