Python tkinter prevent colour box from being called before its supposed to

Question

I have a button that calls a function that asks the user to select a colour. The problem is the colour selection window gets called immediately and the button doesn't work after you input something into the window.

import tkinter as tk
from tkinter.colorchooser import askcolor

main = tk.Tk()

def color_picker(type):
  color = askcolor(title = 'choose ' + type + ' color')
  return color

track_color = tk.Button(main, text='chose track color', command = color_picker('track'))
track_color.pack()

try :
  main()
except TypeError:
  pass

I have no idea why this happens as the function, in theory, is only called when the button is pressed.

Change `command=color_picker("track")` to `command=lambda: color_picker("track")` — TheLizzard, Apr 27 '21 at 13:14
Also why are you returning from the callback? You know that the result will be thrown out — TheLizzard, Apr 27 '21 at 13:15
@TheLizzard im planning on using it later on to colour the line on a matplotlib graph — soap and butter sandwhich, Apr 27 '21 at 13:17
The do that in the function. Everything that you return isn't going to be used. — TheLizzard, Apr 27 '21 at 13:18
@TheLizzard ok i'll set it up to draw the graph in the function. and is there a way i cant limit the amout of colour windows created as ive found that if you spam the button you can create lots of windows — soap and butter sandwhich, Apr 27 '21 at 13:23
Since the color chosen cannot be accessed elsewhere, so what the point of calling it by clicking the button? — acw1668, Apr 27 '21 at 13:28
@acw1668 ngl i dont really know what im doing i'm planning to have a graph that you can change the colour of and im just trying things out this is basically the only solution of could think of im going to tweak the code to have it accessible from else where. — soap and butter sandwhich, Apr 27 '21 at 13:32

Tkinter Lover · Accepted Answer · 2021-04-27T13:29:58.343

1

Since you have parenthesis on your function, it gets called before the color is chosen. To prevent this, use lambda.

So replace line 10 with track_color = tk.Button(main, text='chose track color', command = lambda: color_picker('track')).

Your new code:

import tkinter as tk
from tkinter.colorchooser import askcolor

main = tk.Tk()


def color_picker(type):
  color = askcolor(title = 'choose ' + type + ' color')


track_color = tk.Button(main, text='chose track color', command = lambda: color_picker('track'))
track_color.pack()

main.mainloop()

edited Apr 27 '21 at 13:29

answered Apr 27 '21 at 13:18

Tkinter Lover

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no point of that `return` statement – Matiiss Apr 27 '21 at 13:29
yeah i know i removed it. (I just copied the op's code) – Tkinter Lover Apr 27 '21 at 13:34
@TkinterLover isn't this exactly what I said in the comments bellow the question? – TheLizzard Apr 27 '21 at 13:56
Yeah it looks like it. When I was typing my answer you posted your comment. – Tkinter Lover Apr 27 '21 at 13:58

Python tkinter prevent colour box from being called before its supposed to

1 Answers1