CODEWITHSUNDEEP
pythontkinter
Martinn Roelofse
Martinn Roelofse

Reputation: 433

Python Set Button Text While busy

I'm new to python and I am trying to create a program but I can't even get the basics right. I have a button app that looks like this:

#simple GUI
from tkinter import *
import time

#create the window
root = Tk()

#modify root window
root.title("Button Example")
root.geometry("200x50")

button1state = 0

def start():
    count = 0
    button1["text"] ="Busy!"
    while (count < 5):
        root.after(1000)
        count = count + 1

def button1clicked():
    global button1state
    if button1state == 0:
        start()
        button1["text"] ="On!"
        button1state = 1
    else:
        button1["text"] ="Off!"
        button1state = 0

app = Frame(root)
app.pack()

button1 = Button(app, text ="Off!", command = button1clicked)
button1.pack()

#kick off the event loop
root.mainloop()

Now everything works except it doesn't change the button text to busy while **start()** is called. How can I fix this? Once I've got it working I want to use images to show the user that its OFF ON and BUSY. Please help me

Upvotes: 0

Views: 275

Answers (2)

j_4321
j_4321

Reputation: 16169

You need to force the GUI to update before starting the task:

def start():
    count = 0
    button1.configure(text="Busy!")
    root.update()  # <-- update window
    while (count < 5):
        root.after(1000)
        count = count + 1

But if you don't want your GUI to be frozen while the task is executed, you will need to use a thread as Dedi suggested.

Upvotes: 2

Dedi
Dedi

Reputation: 35

You have to make a thread in order to make you function as a "background event" while your interface is working. Consider using that :

from threading import Thread

and then :

my_thread=Thread(target=start())
my_thread.start()

Where the first "start()" is the name of your function and the second one a call for the thread to begin.

Upvotes: 0

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