How to close a window if tk.Toplevel is in a different file





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how do I close my current window if I can't see the tk.Toplevel(root) from the method I am in?



Here a small example of what i mean:



file1.py:



import Tkinter as tk
import file2
class ExampleMain:
def __init__(self, root):
self.mainFrame = tk.Frame(root)
...
tk.Button(self.mainFrame, command=self.button_pressed)
...

def button_pressed(self):
self.whatever = tk.Toplevel(root)
self.app = file2.ExampleNotMain(self.whatever)

if __name__ == '__main__':
root = tk.Tk()
app = ExampleMain(root)
root.mainloop()


file2.py:



import tkinter as tk
class ExampleNotMain:
def __init__(self, root):
self.frame = tk.Frame(root)
...
tk.Button(self.frame, command=self.close_window)
...

def close_window(self):
=> missing_command_here


In this example I would like to close the second window created (and keep the first).



If all the code is in one file something like



self.whatever.destroy()


would do it. My problem is I can't see the object from the first file with the command being in the second file.



I found something like



execfile("file2.py") 


but I don't like that solution.



Is there a better way to solve my problem?



I would realy apriciate your help.
Thanks in advance.










share|improve this question























  • You code does not show the creation of a second, non-root Toplevel.

    – Terry Jan Reedy
    Nov 16 '18 at 18:59











  • I counted the root as the first window and the non-root window as second. I want to close the non-root window

    – Chris S
    Nov 16 '18 at 19:08













  • I see it now and will answer.

    – Terry Jan Reedy
    Nov 16 '18 at 19:09


















0















how do I close my current window if I can't see the tk.Toplevel(root) from the method I am in?



Here a small example of what i mean:



file1.py:



import Tkinter as tk
import file2
class ExampleMain:
def __init__(self, root):
self.mainFrame = tk.Frame(root)
...
tk.Button(self.mainFrame, command=self.button_pressed)
...

def button_pressed(self):
self.whatever = tk.Toplevel(root)
self.app = file2.ExampleNotMain(self.whatever)

if __name__ == '__main__':
root = tk.Tk()
app = ExampleMain(root)
root.mainloop()


file2.py:



import tkinter as tk
class ExampleNotMain:
def __init__(self, root):
self.frame = tk.Frame(root)
...
tk.Button(self.frame, command=self.close_window)
...

def close_window(self):
=> missing_command_here


In this example I would like to close the second window created (and keep the first).



If all the code is in one file something like



self.whatever.destroy()


would do it. My problem is I can't see the object from the first file with the command being in the second file.



I found something like



execfile("file2.py") 


but I don't like that solution.



Is there a better way to solve my problem?



I would realy apriciate your help.
Thanks in advance.










share|improve this question























  • You code does not show the creation of a second, non-root Toplevel.

    – Terry Jan Reedy
    Nov 16 '18 at 18:59











  • I counted the root as the first window and the non-root window as second. I want to close the non-root window

    – Chris S
    Nov 16 '18 at 19:08













  • I see it now and will answer.

    – Terry Jan Reedy
    Nov 16 '18 at 19:09














0












0








0








how do I close my current window if I can't see the tk.Toplevel(root) from the method I am in?



Here a small example of what i mean:



file1.py:



import Tkinter as tk
import file2
class ExampleMain:
def __init__(self, root):
self.mainFrame = tk.Frame(root)
...
tk.Button(self.mainFrame, command=self.button_pressed)
...

def button_pressed(self):
self.whatever = tk.Toplevel(root)
self.app = file2.ExampleNotMain(self.whatever)

if __name__ == '__main__':
root = tk.Tk()
app = ExampleMain(root)
root.mainloop()


file2.py:



import tkinter as tk
class ExampleNotMain:
def __init__(self, root):
self.frame = tk.Frame(root)
...
tk.Button(self.frame, command=self.close_window)
...

def close_window(self):
=> missing_command_here


In this example I would like to close the second window created (and keep the first).



If all the code is in one file something like



self.whatever.destroy()


would do it. My problem is I can't see the object from the first file with the command being in the second file.



I found something like



execfile("file2.py") 


but I don't like that solution.



Is there a better way to solve my problem?



I would realy apriciate your help.
Thanks in advance.










share|improve this question














how do I close my current window if I can't see the tk.Toplevel(root) from the method I am in?



Here a small example of what i mean:



file1.py:



import Tkinter as tk
import file2
class ExampleMain:
def __init__(self, root):
self.mainFrame = tk.Frame(root)
...
tk.Button(self.mainFrame, command=self.button_pressed)
...

def button_pressed(self):
self.whatever = tk.Toplevel(root)
self.app = file2.ExampleNotMain(self.whatever)

if __name__ == '__main__':
root = tk.Tk()
app = ExampleMain(root)
root.mainloop()


file2.py:



import tkinter as tk
class ExampleNotMain:
def __init__(self, root):
self.frame = tk.Frame(root)
...
tk.Button(self.frame, command=self.close_window)
...

def close_window(self):
=> missing_command_here


In this example I would like to close the second window created (and keep the first).



If all the code is in one file something like



self.whatever.destroy()


would do it. My problem is I can't see the object from the first file with the command being in the second file.



I found something like



execfile("file2.py") 


but I don't like that solution.



Is there a better way to solve my problem?



I would realy apriciate your help.
Thanks in advance.







python tkinter






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 16 '18 at 18:36









Chris SChris S

182




182













  • You code does not show the creation of a second, non-root Toplevel.

    – Terry Jan Reedy
    Nov 16 '18 at 18:59











  • I counted the root as the first window and the non-root window as second. I want to close the non-root window

    – Chris S
    Nov 16 '18 at 19:08













  • I see it now and will answer.

    – Terry Jan Reedy
    Nov 16 '18 at 19:09



















  • You code does not show the creation of a second, non-root Toplevel.

    – Terry Jan Reedy
    Nov 16 '18 at 18:59











  • I counted the root as the first window and the non-root window as second. I want to close the non-root window

    – Chris S
    Nov 16 '18 at 19:08













  • I see it now and will answer.

    – Terry Jan Reedy
    Nov 16 '18 at 19:09

















You code does not show the creation of a second, non-root Toplevel.

– Terry Jan Reedy
Nov 16 '18 at 18:59





You code does not show the creation of a second, non-root Toplevel.

– Terry Jan Reedy
Nov 16 '18 at 18:59













I counted the root as the first window and the non-root window as second. I want to close the non-root window

– Chris S
Nov 16 '18 at 19:08







I counted the root as the first window and the non-root window as second. I want to close the non-root window

– Chris S
Nov 16 '18 at 19:08















I see it now and will answer.

– Terry Jan Reedy
Nov 16 '18 at 19:09





I see it now and will answer.

– Terry Jan Reedy
Nov 16 '18 at 19:09












1 Answer
1






active

oldest

votes


















0














When you create ExampleNotMain, you pass the 2nd Toplevel self.whatever. In ExampleNotMain.__init__, it gets bound to root. (master or parent would be a better parameter name). In __init__, add self.top = root(or whatever you call the passed-in toplevel). Inclose_window, addself.top.destroy()`.






share|improve this answer
























  • It works. Thanks for your help.

    – Chris S
    Nov 16 '18 at 19:22













  • Then please accept and/or upvote the answer.

    – Terry Jan Reedy
    Nov 16 '18 at 19:31












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0














When you create ExampleNotMain, you pass the 2nd Toplevel self.whatever. In ExampleNotMain.__init__, it gets bound to root. (master or parent would be a better parameter name). In __init__, add self.top = root(or whatever you call the passed-in toplevel). Inclose_window, addself.top.destroy()`.






share|improve this answer
























  • It works. Thanks for your help.

    – Chris S
    Nov 16 '18 at 19:22













  • Then please accept and/or upvote the answer.

    – Terry Jan Reedy
    Nov 16 '18 at 19:31
















0














When you create ExampleNotMain, you pass the 2nd Toplevel self.whatever. In ExampleNotMain.__init__, it gets bound to root. (master or parent would be a better parameter name). In __init__, add self.top = root(or whatever you call the passed-in toplevel). Inclose_window, addself.top.destroy()`.






share|improve this answer
























  • It works. Thanks for your help.

    – Chris S
    Nov 16 '18 at 19:22













  • Then please accept and/or upvote the answer.

    – Terry Jan Reedy
    Nov 16 '18 at 19:31














0












0








0







When you create ExampleNotMain, you pass the 2nd Toplevel self.whatever. In ExampleNotMain.__init__, it gets bound to root. (master or parent would be a better parameter name). In __init__, add self.top = root(or whatever you call the passed-in toplevel). Inclose_window, addself.top.destroy()`.






share|improve this answer













When you create ExampleNotMain, you pass the 2nd Toplevel self.whatever. In ExampleNotMain.__init__, it gets bound to root. (master or parent would be a better parameter name). In __init__, add self.top = root(or whatever you call the passed-in toplevel). Inclose_window, addself.top.destroy()`.







share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 16 '18 at 19:17









Terry Jan ReedyTerry Jan Reedy

12.3k12140




12.3k12140













  • It works. Thanks for your help.

    – Chris S
    Nov 16 '18 at 19:22













  • Then please accept and/or upvote the answer.

    – Terry Jan Reedy
    Nov 16 '18 at 19:31



















  • It works. Thanks for your help.

    – Chris S
    Nov 16 '18 at 19:22













  • Then please accept and/or upvote the answer.

    – Terry Jan Reedy
    Nov 16 '18 at 19:31

















It works. Thanks for your help.

– Chris S
Nov 16 '18 at 19:22







It works. Thanks for your help.

– Chris S
Nov 16 '18 at 19:22















Then please accept and/or upvote the answer.

– Terry Jan Reedy
Nov 16 '18 at 19:31





Then please accept and/or upvote the answer.

– Terry Jan Reedy
Nov 16 '18 at 19:31




















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