蟒蛇，Tkinter的，如何摧毁标签在与循环两种不同的标签？(python, tkinter, ho

感谢@ Miraj50，帮助我与.destroy（） Tkinter的：如何编写一个for循环破坏标签列表？下一阶段，我试图摧毁这两个选项卡的标签。我知道如何共享同一份名单，不同的标签，但我不知道如何“链接”他们的标签。用我有限的知识，我试过

def remove(self, name):
    for name in tabs[name]:
        for employee in labelemployee:
            labelemployee[employee].destroy()

它给我这个错误：

TypeError: remove() missing 1 required positional argument: 'name'

然后，我尝试

"for name in tabs["Requests"]"

只是为了看看是否可行与否。它仍然有同样的错误。如果有人可以帮助我解决这个驼背并清除我的困惑，请。下面是完整的代码：

import tkinter as tk
from tkinter import ttk

labelemployee={}
upper_tabs = ["Final", "Requests", "Control"]
tabs = {}

class Application(ttk.Frame): #inherent from frame.

    def __init__(self, parent):
        tk.Frame.__init__(self, parent, bg="LightBlue4")
        self.parent = parent
        self.Employees = ["A", "B", "C", "D"]
        self.pack(fill=tk.BOTH, expand=1)
        self.GUI()

    def GUI(self): #the function that runs all the GUI functions.
        self.create_tabs()
        self.buttons("Control")
        for name in upper_tabs[:2]:
        self.create_grid(name)
        self.add_left_names("Requests")
        self.add_left_names("Final")

    def create_tabs(self):
        self.tabControl = ttk.Notebook(self, width=1000, height=400)
        for name in upper_tabs:
            self.tab=tk.Frame(self.tabControl, bg='thistle')
            self.tabControl.add(self.tab, text=name)
            tabs[name] = self.tab
            self.tabControl.grid(row=0, column=0, sticky='nsew')   

    def create_grid(self, name):
        for i in range (7):
            for j in range(7):
                self.label = tk.Label(tabs[name], relief="ridge", 
                     width=12, height=3)
                self.label.grid(row=i, column=j, sticky='nsew')

    def buttons(self, name):
        self.button=tk.Button(tabs[name], text="Clear", bg="salmon",   
            command = self.remove)
        self.button.pack()

    def add_left_names(self, name):
       #--------add in name labels on the side--------------        
        i=2
        for employee in self.Employees:
            self.label=tk.Label(tabs[name], text=employee ,  fg="red", 
               bg="snow")
            self.label.grid(row=i,column=0)
            labelemployee[employee]=self.label
            i +=1

    **def remove(self, name):
        for name in tabs[name]:
            for employee in labelemployee:
            labelemployee[employee].destroy()**

def main():
    root = tk.Tk()
    root.title("class basic window")
    root.geometry("1000x500")
    root.config(background="LightBlue4")
    app = Application(root)
    root.mainloop()

if __name__ == '__main__':
    main()

首先，你需要保持所有标签的轨道。这里只是add_left_names分配标签将覆盖旧标签。所以，我存储在一个列表，它的价值是给员工关键的标签。现在，在删除功能，你只需要在所有这些标签在迭代和labelemployee摧毁他们。

import tkinter as tk
from tkinter import ttk
from collections import defaultdict

labelemployee=defaultdict(list)
upper_tabs = ["Final", "Requests", "Control"]
tabs = {}

class Application(ttk.Frame): #inherent from frame.

    def __init__(self, parent):
        tk.Frame.__init__(self, parent, bg="LightBlue4")
        self.parent = parent
        self.Employees = ["A", "B", "C", "D"]
        self.pack(fill=tk.BOTH, expand=1)
        self.GUI()

    def GUI(self): #the function that runs all the GUI functions.
        self.create_tabs()
        self.buttons("Control")
        for name in upper_tabs[:2]:
            self.create_grid(name)
        self.add_left_names("Requests")
        self.add_left_names("Final")

    def create_tabs(self):
        self.tabControl = ttk.Notebook(self, width=1000, height=400)
        for name in upper_tabs:
            self.tab=tk.Frame(self.tabControl, bg='thistle')
            self.tabControl.add(self.tab, text=name)
            tabs[name] = self.tab
            self.tabControl.grid(row=0, column=0, sticky='nsew')   

    def create_grid(self, name):
        for i in range (7):
            for j in range(7):
                self.label = tk.Label(tabs[name], relief="ridge", width=12, height=3)
                self.label.grid(row=i, column=j, sticky='nsew')

    def buttons(self, name):
        self.button=tk.Button(tabs[name], text="Clear", bg="salmon", command=self.remove)
        self.button.pack()

    def add_left_names(self, name):
       #--------add in name labels on the side--------------        
        i=2
        for employee in self.Employees:
            self.label=tk.Label(tabs[name], text=employee, fg="red", bg="snow")
            self.label.grid(row=i,column=0)
            labelemployee[employee].append(self.label)
            i +=1

    def remove(self):
        for employee in labelemployee:
            for label in labelemployee[employee]:
                label.destroy()

def main():
    root = tk.Tk()
    root.title("class basic window")
    root.geometry("1000x500")
    root.config(background="LightBlue4")
    app = Application(root)
    root.mainloop()

if __name__ == '__main__':
    main()