python，tkinter，如何使用for循环销毁两个不同选项卡中的标签？

Question

感谢 @Miraj50 帮助我完成 .destroy() tkinter: how to write a for loop to destroy a list of labels? 在下一阶段，我正在尝试销毁两个选项卡中的标签。我知道如何将相同的列表共享到不同的选项卡，但我不知道如何将它们“链接”到选项卡。以我有限的知识，我尝试了

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

它给了我这个错误：

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

然后我尝试了

"for name in tabs["Requests"]" 

只是看看它是否有效。它仍然有同样的错误。如果有人可以帮助我解决这个问题并消除我的困惑，请。以下是完整代码：

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()

Answer 1

首先，您需要跟踪所有标签。这里只分配 add_left_names 中的标签将覆盖之前的标签。所以我将标签存储在一个列表中，该列表是员工 key 的 value。现在在 remove 函数中，您只需要遍历 labelemployee 中的所有这些标签并销毁它们。

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()