14

I'm having trouble with a function that shows an image for two seconds on screen, and then is destroyed. When the program runs the functions initial call procedurely works fine, but if the function is then called via a button built in tkinter I get an error.

appcwd = os.getcwd()
user32 = ctypes.windll.user32
screensize = user32.GetSystemMetrics(0), user32.GetSystemMetrics(1)
size = str(screensize[0])+'x'+str(screensize[1])

def wlcm_scrn(event=None):
    def destroy_wlcm(event=None):
        wlcm_scrn.destroy()
    global appcwd
    global screensize
    wlcm_scrn = tkinter.Tk()
    file=appcwd+"\\Run_Files\\splash.gif"
    splsh_img = tkinter.PhotoImage(file=file) 
    splosh = tkinter.Label(wlcm_scrn,image=splsh_img)
    wlcmh = splsh_img.height()/2
    wlcmw = splsh_img.width()/2
    splosh.pack()
    wlcm_scrn.config(bg='black')
    wlcm_scrn.overrideredirect(True)
    wlcm_scrn.bind("<Escape>",destroy_wlcm)
    wlxym = '+'+str(int((screensize[0]/2)-wlcmw))+'+'+str(int((screensize[1]/2)-wlcmh))
    wlcm_scrn.geometry(wlxym)
    wlcm_scrn.wm_attributes("-topmost", 1)
    wlcm_scrn.after(2000,destroy_wlcm)
    wlcm_scrn.mainloop()

wlcm_scrn() #Call through procedure.

Button that calls the function.

view_img = tkinter.Button(cfrm,text='Show splash image',command=wlcm_scrn)

Error message when called through button command.

Exception in Tkinter callback
Traceback (most recent call last):
  File "C:\Python33\lib\tkinter\__init__.py", line 1475, in __call__
    return self.func(*args)
  File "C:\Python33\POS_Solution\Rattle_Hum_POS.py", line 1755, in run_wlcm_scrn
    wlcm_scrn()
  File "C:\Python33\POS_Solution\Rattle_Hum_POS.py", line 34, in wlcm_scrn
    splosh = tkinter.Label(wlcm_scrn,image=splsh_img)
  File "C:\Python33\lib\tkinter\__init__.py", line 2596, in __init__
    Widget.__init__(self, master, 'label', cnf, kw)
  File "C:\Python33\lib\tkinter\__init__.py", line 2075, in __init__
    (widgetName, self._w) + extra + self._options(cnf))
_tkinter.TclError: image "pyimage3" doesn't exist

What is "pyimage3" and why doesn't it exist? Any help would be apprecaited. Thanks.

I_do_python
  • 1,366
  • 4
  • 16
  • 31
  • have a look here http://www.daniweb.com/software-development/python/threads/154237/tkinter-problem – xor Nov 27 '13 at 19:34
  • I don't think it's a problem with the file path, as I'm using os.getcwd(), and it works first time around. – I_do_python Nov 28 '13 at 05:58
  • Also the bit by Bennett_1 who states "...the file reference had been incorrect on a previous run." I don't think this is applicable as it worked on the first run, and the file reference doesn't change. – I_do_python Nov 28 '13 at 06:00
  • I can run it multiple times procedurely and it works everytime, so there must be something about calling it as an event on a tkinter button that is causing the issue. – I_do_python Nov 28 '13 at 06:09

6 Answers6

37

I found the issue so figured I'd answer myself for anyone who has this issue in the future.

When the wlcm_scrn runs procedurely it is the only window that exists at that point in time, and so it can use tkinter.Tk(). The error arises because the button that calls the function is itself sitting in an active window that is also running as Tkinter.Tk(). So when Python/Tkinter tries to build wlcm_scrn from the button, it's essentially trying to create two windows under root and falling over.

The solution:

Changing line...

wlcm_scrn = tkinter.Tk()

to this...

wlcm_scrn = tkinter.Toplevel()

...stops the error, and the image shows.

I personally am going to have two instances of the function. One called procedurely under Tk(), and one called within the application under TopLevel().

I_do_python
  • 1,366
  • 4
  • 16
  • 31
7

The PhotoImage method creates an image for the first TK () instance created. Thus, it seems to have solved to inherit TK () instance by replacing TopLevel ().

This can be solved by specifying the master of the Tk () instance as the option of PhotoImage.

I think this should be changed.:

splsh_img = tkinter.PhotoImage(file=file,master=wlcm_scrn)
Takahiro Funahashi
  • 272
  • 4
  • 10
1

I had been getting the same error, and the above methods work well to fix it, but i also found that if you aren't going to be changing all of this, then you can restart the kernel (in jupyter notebook) or restart your python interpreter

And that should make it work the next time. I am not entirely sure how and why this works tho, but it does and is a simple solution for the time being

Krishnaraj PT
  • 138
  • 1
  • 11
  • be more specific while answering the question. And never guess the output and post when you are 100% sure otherwise can leave a comment. – Suryaveer Singh Sep 05 '20 at 08:50
0

Maybe not for this exact case, but for a similar one, I found that the customary 

if __name__ == '__main__':
    wlcm_scrn()  #Call through procedure.

solves the problem as well. This seems to kill the active window and to create a new one each time you re-call the same module.

SeF
  • 3,864
  • 2
  • 28
  • 41
0
from tkinter import *
import tkinter as tk
window = tk.Tk()

window.title('Learn')
window.geometry("500x500")

text = Label(window, text = 'label here')
text.pack()

photo = PhotoImage(file='C:\\Users\\user\\Documents\\image\\image_2.png')
labelphoto = Label(window , image = photo)
labelphoto.pack()

window.mainloop()enter code here
user19647767
  • 1
  • 1
    Your answer could be improved with additional supporting information. Please [edit] to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers [in the help center](/help/how-to-answer). – Community Jul 31 '22 at 19:28
0

this is one example of code

Root =Tk()
window = Root.Toplevel()

photo =PhotoImage(file='C:\\Users\\user\\Documents\\image\\image_2.png', master= window)

labelphoto = Label(window , image = photo)
labelphoto.pack()

I use "master" in photoImage to specify parent screen . Thank you

toyota Supra
  • 3,181
  • 4
  • 15
  • 19
KevinKefra1
  • 1