Tkinter error on a label object

Question

First good afternoon friends, I am developing a POO application in python and my object returned the following error:

  File "codigo_principal.py", line 37, in <module>
    menu=Autenticar(["Roboto",10],"LavenderBlush",[1,40],('400x150'),"MENU",True)
  File "/home/bart/teste/janela.py", line 29, in __init__
    self.nome=Criar_Label(self.root,"NOME",xy[0],xy[1],fonte,fundo)
  File "/home/bart/teste/criar_widgets.py", line 6, in __init__
    self.label=Label(root)
  File "/usr/lib/python3.6/tkinter/__init__.py", line 2766, in __init__
    Widget.__init__(self, master, 'label', cnf, kw)
  File "/usr/lib/python3.6/tkinter/__init__.py", line 2299, in __init__
    (widgetName, self._w) + extra + self._options(cnf))
_tkinter.TclError: can't invoke "label" command: application has been destroyed

I searched the Python documentation, more specifically Tkinter and realized that this error usually occurs when there are window problems when destroying the previous one to create a new one. However in my case I’m not trying to do it, my code so far only has a window!

Follow the files that form the code: 1-main code.py

from janela import *
menu=Autenticar(["Roboto",10],"LavenderBlush",[1,40],'400x150'),"MENU",True)

2-py

from tkinter import *
from criar_widgets import *

class Janelas:
   def __init__(self,fonte,fundo,xy,tela,titulo,original):
     #Caso seja a janela mãe
     if original:
          #Esta é a variável da janela mãe
          self.root = Tk()
     #Caso seja uma janela filha

     else:
          #Esta é a variável da janela filha
          self.root= Toplevel()

     #Esta é a cor de fundo da janela
     self.root["bg"]=fundo
     #Este é o título
     self.titulo=Criar_Label_Titulo(self.root,titulo,fonte,fundo)
     #Este é o tamanho da janela
     self.root.geometry(tela)
     self.root.mainloop()

class Autenticar(Janelas):
    def __init__(self,fonte,fundo,xy,tela,titulo,original):

        super().__init__(fonte,fundo,xy,tela,titulo,original)

        self.nome=Criar_Label(self.root,"NOME",xy[0],xy[1],fonte,fundo)
        self.entrar_nome=Criar_Entry(self.root,xy[0]+50,xy[1],fonte,fundo,False)  

        self.senha=Criar_Label(self.root,"SENHA",xy[0],xy[1]+30,fonte,fundo)
        self.entrar_senha=Criar_Entry(self.root,xy[0]+50,xy[1]+30,fonte,fundo,True)

        self.botao=Criar_Button(self.root,"INSERIR NOME",fonte,fundo,lambda:login(self.root,usuario,senha))

    def login(self,usuario,senha):
        if usuario == "" and senha == "":
           pass
            #self.root.destroy()

3-create widgets.py

from tkinter import *

class Criar_Label:
    def __init__(self,root,texto,abcissa,ordenada,fonte,fundo):
         self.label=Label(root)
         self.label["text"]=texto
         self.label["font"]=fonte
         self.label["bg"]=fundo
         self.label.place(x=abcissa,y=ordenada)

#Será que deveríamos fazer polimorfismo aqui e fazer da classe Criar_Label a classe mãe?
class Criar_Label_Titulo:
    def __init__(self,root2,texto,fonte,fundo):
         self.titulo=Label(root2)
         self.titulo["text"]=texto
         self.titulo["font"]=fonte
         self.titulo["bg"]=fundo
         self.titulo.pack()

class Criar_Entry:
    def __init__(self,root2,abcissa,ordenada,fonte,fundo,segredo):
         self.entry=Entry(root2,font=fonte)
         self.entry.place(x=abcissa,y=ordenada)
         if segredo:
             self.entry["show"]="*"

class Criar_Button:
    def __init__(self,root2,texto,fonte,fundo,comando):
         self.button=Button(root2)
         self.button["text"]=texto
         self.button["font"]=fonte
         self.button["bg"]=fundo
         self.button["command"]=comando

         self.button.pack(side=BOTTOM)

Until you open the window and put a widget label,but hangs and when closes gives the error. The documentation I have studied to come to the conclusion that it is an error to open or close a window are the following pages:

1- The stack overflow page itself

https://stackoverflow.com/questions/29655219/whats-the-difference-between-tkinters-tk-and-toplevel-classes

2- O Reddit

https://www.reddit.com/r/learnpython/comments/cx4lu9/cant_invoke_label_error_using_tkinter_on_python_36/

I do not know if I misunderstood the documentation or not, but the error exists and I do not know correct it.

Grateful

Answer 1

solved the problem by reversing the inheritance

in your code you create

class Janelas
class Autenticar(Janelas)

if you reverse to

class Autenticar
class Janelas(Autenticar)

and correct the call of super(), the code runs showing the label and entry smoothly

Here’s how I organize my apps, you can see that the settings are in other classes

from gui import GUI
from gerador import Gerador
from conf import Setup
from mapas import ZipA

class App(GUI, Gerador, Setup, ZipA):
    VER = 'Maio/2019'

app = App()

GUI is the class that takes care of the design

Setup is the class that connects functionality with design

---

seen as this the code in git, I thought in a slightly different media

class Autenticar:
    def __init__(self,fonte,fundo,xy,tela,titulo,original):
        self.nome=Criar_Label(self.root,"NOME",xy[0],xy[1],fonte,fundo)
        self.entrar_nome=Criar_Entry(self.root,xy[0]+50,xy[1],fonte,fundo,False)  

        self.senha=Criar_Label(self.root,"SENHA",xy[0],xy[1]+30,fonte,fundo)
        self.entrar_senha=Criar_Entry(self.root,xy[0]+50,xy[1]+30,fonte,fundo,True)

        self.botao=Criar_Button(self.root,"INSERIR NOME",fonte,fundo,lambda:login(self.root,usuario,senha))

    def login(self,usuario,senha):
        if usuario == "" and senha == "": 
           pass
            #self.root.destroy()

class Janelas(Autenticar):
   def __init__(self,fonte,fundo,xy,tela,titulo,original):
     if original:
          self.root = Tk()

     else:
          self.root= Toplevel()
     super().__init__(fonte,fundo,xy,tela,titulo,original)

     self.root["bg"]=fundo
     self.titulo=Criar_Label_Titulo(self.root,titulo,fonte,fundo)
     self.root.geometry(tela)
     self.root.mainloop()


menu=Janelas(["Roboto",10],"LavenderBlush",[1,40],'400x150',"MENU",True)