问题描述
我正在尝试从网格中删除这些元素.通过逐一写出,我可以删除所有这些文件.然后,我编写了一个for循环以使其可扩展,然后遇到此错误消息.
I am trying to remove these elements from the grid. I was able to delete all of them by writing out one by one. I then wrote a for loop to make it expandable, then I run into this error message.
"employee.destroy()
AttributeError: 'str' object has no attribute 'destroy'"
这是一个更大程序的一部分,但是我可以尽最大努力解决核心问题,这是我的代码:
This is part of a bigger program, but as much as I can reduce to the core problem, here is my code:
import tkinter as tk
from tkinter import ttk
labelemployee={}
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.buttons()
self.create_grid()
self.add_left_names()
def remove(self):
#labelemployee["A"].destroy()
#labelemployee["B"].destroy()
#labelemployee["C"].destroy()
#labelemployee["D"].destroy()
for employee in labelemployee:
employee.destroy()
def create_grid(self):
for i in range (7):
for j in range(7):
self.label = tk.Label(self, relief="ridge", width=12,
height=3)
self.label.grid(row=i, column=j, sticky='nsew')
def buttons(self):
self.button=tk.Button(self, text="Clear", bg="salmon", command
= self.remove)
self.button.grid(row=7, column=6, sticky='e')
def add_left_names(self):
#--------add in name labels on the side--------------
i=2
for employee in self.Employees:
self.label=tk.Label(self, text=employee , fg="red",
bg="snow")
self.label.grid(row=i,column=0)
labelemployee[employee]=self.label
i +=1
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()
请帮助我.我认为问题是我的for循环存储了一个列表,并且我有一个字典.因此,我想我不知道如何破坏字典中的标签.
Please help me. I think the problem is my for loop is stored a list, and I have a dictionary. So, then I guess i don't know how to destroy labels in a dictionary.
推荐答案
您已经在这些评论中找到了问题!
You have already figured out the issue in those comments !
您已经知道labelemployee
是字典,因此在默认情况下对其进行迭代将为您提供字典的键.因此,employee
将是诸如 A , B ...之类的字符串.销毁字符串对象显然会给您一个错误.您需要销毁相应的tkinter小部件.因此,您应该在for循环中将employee.destroy()
替换为labelemployee[employee].destroy()
.
You already know that labelemployee
is a dictionary, so iterating over it will give you, by default, the keys of the dictionary. So employee
will be strings like A, B... and so on. And destroying a string object will obviously give you an error. You need to destroy the corresponding tkinter widget. So for that, you should replace employee.destroy()
with labelemployee[employee].destroy()
in the for loop.
这篇关于tkinter:如何编写一个for循环来销毁标签列表?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!