类型错误:'NoneType'对象不支持项目分配?

本文介绍了类型错误:'NoneType'对象不支持项目分配?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

限时送ChatGPT账号..

所以我用文件选择"和去"按钮启动一个根屏幕.转到按钮已禁用，我想在选择文件后将其激活.当我选择文件时，go 应该变为活动状态，而是出现此错误TypeError:'NoneType' 对象不支持项目分配"这是一些示例代码

So I start a root screen with a "file select" and a "go" button. The go button is disabled and I want to make it active after the file has been selected. When I select the file go should become active but instead this error "TypeError: 'NoneType' object does not support item assignment" Here is some sample code

import Tkinter
import tkFileDialog

def chooseDir():
    dir1=tkFileDialog.askopenfilename(parent=root, title='choose file path')
    go['state']=Tkinter.ACTIVE
root=Tkinter.Tk()
global go
go=Tkinter.Button(text='file location',command=chooseDir,state=Tkinter.DISABLED).pack()
root.mainloop()

推荐答案

这一行:

go=Tkinter.Button(text='file location',command=chooseDir,state=Tkinter.DISABLED).pack()

正在创建一个临时对象，然后对其调用 pack().pack 方法返回 None，所以 go 被分配了 None.

is creating a temporary object, then calling pack() on it. The pack method returns None, so go is assigned None.

删除 .pack() 然后 go 将成为 Button 对象.然后调用go.pack().

Remove the .pack() then go will be the Button object. Then call go.pack().

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