tkinter子窗口按钮来改变列表框的内容
这个问题是可以解决的。

• 将cLabelChange()函数移动到child()函数内部。
• 将cLabelChange()函数的Cbtn和pack移动到底部。

仅此而已
片段：

import tkinter as tk
root = tk.Tk()
root.title("Root")
root.geometry("300x300")
 

def child(): #Create child window
    child = tk.Toplevel()
    child.title("Child Window")
    child.geometry("350x350+100+100")
         
    lb1 = tk.Label(child, text = "Child Label")
    lb1.pack()
         
    def cLabelChange():
        lb1.configure(text = "I changed this text") #THIS IS WHERE THE ERROR OCCURS
        child.update()

    Cbtn = tk.Button(child, text = "Change Child Label text", command = cLabelChange)
    Cbtn.pack()    

btn = tk.Button(root, text = "Open Child.", command = child)
btn.pack()

root.mainloop()

截图：

这可能是您的问题的替代解决方案。将lb1作为参数传递给labelChange()函数，这样就可以在函数内部访问标签小部件，如下所示。要做到这一点，你必须使用lambda函数传递参数。

import tkinter as tk
root = tk.Tk()
root.geometry("300x300")
root.title("Root")

def child(): #Create child window
    child = tk.Toplevel()
    child.geometry("350x350+100+100")
    child.title("Child Window")
    
    lb1 = tk.Label(child, text = "Child Label")
    lb1.pack()
    
    Cbtn = tk.Button(child, text = "Change Child Label text", command = lambda:cLabelChange(lb1))
    Cbtn.pack()

def cLabelChange(lb1):
    lb1.configure(text = "I changed this text") #THIS IS WHERE THE ERROR OCCURS

btn = tk.Button(root, text = "Open Child.", command = child)
btn.pack()

root.mainloop()