我正在尝试插入名为'变量A'的变量。来自' Class_A'将类分类到另一个名为' App(tk)'的类的列表框中。任何人都可以帮助我吗?
然而,当我使用我的代码时,以及在运行'运行'按钮,它打开一个新窗口(与主窗口相同),而不是插入变量A'进入'积极'现有窗口中的列表框。 我也收到错误:AttributeError:' _tkinter.tkapp'对象没有属性' myListbox_01'
以下是代码:
from tkinter import *
import tkinter.messagebox as tkMessageBox
import tkinter.filedialog as tkFileDialog
class class_A(object):
def __init__(self, isON = False):
self.isOn = False
def turnOn(self):
self.isOn = True
tkMessageBox.showinfo(title = 'On', message = 'It is on')
x = 'favorable'
if x == 'favorable':
temp_App = App()
temp_App.myListbox_01.insert(END, 'Variable A')
else:
temp_App.myListbox_02.insert(END, 'Variable A')
def turnOff(self):
self.isOn = False
tkMessageBox.showinfo(title = 'Off', message = 'It is off')
class App(Tk):
def __init__(self):
Tk.__init__(self)
def toggle():
if button.config('text')[-1] == 'Run':
A = class_A()
A.turnOn()
button.config(text='Stop')
else:
button.config(text='Run')
A = class_A()
A.turnOff()
Frame_0 = Frame(self, bg = 'Black', borderwidth = 2, relief = GROOVE)
Frame_0.pack(side = TOP, padx = 10, pady = 2.5)
# Positive
Frame_title01 = Frame(Frame_0, bg="white", height = 10, width = 300, borderwidth = 2, relief=GROOVE)
Frame_title01.grid(row = 0, column = 0, padx=5, pady=5)
Label(Frame_title01, text="positive").pack(padx=2, pady=2)
Frame_01 = Frame(Frame_0, bg="white", height = 200, width = 300, borderwidth = 2, relief=GROOVE)
Frame_01.grid(row = 1, column = 0, padx=5, pady=5)
myListbox_01 = Listbox(Frame_01, bg = 'white', width = 15, height = 10, font = ('times', 14), borderwidth=0)
myListbox_01.grid(row = 0, column = 0, padx = 80, pady = 5)
# Negative
Frame_title02 = Frame(Frame_0, bg="white", height = 10, width = 300, borderwidth = 2, relief=GROOVE)
Frame_title02.grid(row = 0, column = 1, padx=2, pady=2)
Label(Frame_title02, text="Negative").pack(padx=2, pady=2)
Frame_02 = Frame(Frame_0, bg="white", height = 200, width = 300, borderwidth = 2, relief=GROOVE)
Frame_02.grid(row = 1, column = 1, padx=5, pady=5)
myListbox_02 = Listbox(Frame_02, bg = 'white', width = 15, height = 10, font = ('times', 14), borderwidth=0)
myListbox_02.grid(row = 0, column = 0, padx = 80, pady = 5)
# Button
Frame_1 = Frame(self, bg = 'white', borderwidth = 2, relief = FLAT)
Frame_1.pack(side = TOP)
button = Button(Frame_1, text = 'Run', command = toggle)
button.pack(pady = 10)
if __name__ == "__main__":
app = App()
app.geometry("800x300+51+51")
app.title("GUI")
app.mainloop()
答案 0 :(得分:1)
您正在获取一个新窗口,因为您的class_A
实例在App
中创建了一个新的turnOn
实例。您可能希望在某个时刻将现有的App
传递给类(要么是构造函数,要么传递给turnOn
本身。)
这是一个快速而肮脏的修复方法。更好的版本可能会将app
值保留在class_A
中的实例变量中,而不是仅将其传递给turnOn
(您可能还希望App
保留一个class_A
的实例,而不是每次按下按钮时都创建一个新实例:
class class_A(object):
def __init__(self, isON = False):
self.isOn = False
def turnOn(self, app): # new app arg!!!
self.isOn = True
tkMessageBox.showinfo(title = 'On', message = 'It is on')
x = 'favorable'
if x == 'favorable':
app.myListbox_01.insert(END, 'Variable A') # use new arg here
else:
app.myListbox_02.insert(END, 'Variable A') # and here
def turnOff(self):
self.isOn = False
tkMessageBox.showinfo(title = 'Off', message = 'It is off')
class App(Tk):
def __init__(self):
Tk.__init__(self)
def toggle():
if button.config('text')[-1] == 'Run':
A = class_A()
A.turnOn(self) # pass self as app arg!
button.config(text='Stop')
else:
button.config(text='Run')
A = class_A()
A.turnOff()
# ...