类型错误:'NoneType'对象不支持项目分配? [英] TypeError: 'NoneType' object does not support item assignment?

问题描述

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().

这篇关于类型错误:'NoneType'对象不支持项目分配?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IT屋！