CODEWITHSUNDEEP
pythontkintermenupython-os
Matthew
Matthew

Reputation: 37

How do I make this python GUI run this executable?

I'm trying to make a basic gui menu option that will run my stated exe (hurl), but it won't. Here is my code:

from tkinter import *
import os
root = Tk()
root.geometry('350x150')
root.title("hurler")
photo = PhotoImage(file = "Logo_Image.png")
root.iconphoto(False, photo)

entry_text = Label(text = "Number of posts you wish to automate (between 1-12) * ")

entry_text.place(x = 15, y = 10)

num = StringVar()

time_entry = Entry(textvariable = num, width = "10")

time_entry.place(x = 15, y = 40)

def action():
    if num == '1':
        os.system(".\hurl\hurl.exe")

register = Button(root,text = "Make", width = "10", height = "2", command = action, bg = "lightblue")
register.place(x = 15, y = 70)

root.mainloop()

I'm not the most experienced at this, so any feedback helps.

How do I get option 1 to run this exe? Thanks.

Upvotes: 0

Views: 67

Answers (1)

Matiiss
Matiiss

Reputation: 6156

so these are the improvements, should work now:

from tkinter import *
import os
root = Tk()
root.geometry('350x150')
root.title("hurler")
# photo = PhotoImage(file = "Logo_Image.png")
# root.iconphoto(False, photo)

entry_text = Label(text = "Number of posts you wish to automate (between 1-12) * ")

entry_text.place(x = 15, y = 10)

num = StringVar()

time_entry = Entry(textvariable = num, width = "10")

time_entry.place(x = 15, y = 40)

def action():
    global num
    num = num.get()
    if num == '1':
        os.startfile('https://www.google.com')
    num = StringVar()

register = Button(root,text = "Make", width = "10", height = "2", command = action, bg = "lightblue")
register.place(x = 15, y = 70)

root.mainloop()

the changes were made here:

def action():
    global num
    num = num.get()
    if num == '1':
        os.startfile('https://www.google.com')
    num = StringVar()

also should work with the os.system thingy

Upvotes: 1

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