TypeError：“NoneType”对象在 Tkinter 中不可调用

Question

我是 tkinter 新手并尝试了一个简单的 Tkinter GUI，但出现错误：

TypeError：“NoneType”对象不可调用

这是我的代码：

from tkinter import *

from tkinter import filedialog

import pandas as pd

import pyodbc

from sqlalchemy import create_engine

import urllib

master = Tk()

master.title("Demo GUI")

master.geometry("900x400+150+150")

master.resizable(0,0)

def browse_file():

    global file_path

    global data_frame

    file_path = filedialog.askopenfilename(title = "Choose the file to upload")

    data_frame = pd.read_excel(file_path)

Label = Label(master,text="Choose the file to upload").grid(row=0)

Button = Button(master,text='Browse',command = browse_file).grid(row=0,column=1,pady=4)

Label_1 = Label(master,text="The file selected: "+file_path).grid(row=1,column=0)

master.mainloop()

我得到的错误是，

TypeError                                 Traceback (most recent call last)

<ipython-input-29-451372edd65a> in <module>

     25 Button = Button(master,text='Browse',command = browse_file).grid(row=0,column=1,pady=4)

     26 

---> 27 Label1 = Label(master,text="The file Choosen: "+file_path).grid(row=1,column=0)

     28 

     29 

TypeError: 'NoneType' object is not callable

Answer 1

'NoneType' 对象不可调用错误是由将对象放置在定义的位置引起的。所以而不是Label_1&nbsp;=&nbsp;Label(master,text="The&nbsp;file&nbsp;selected:&nbsp;"+file_path).grid(row=1,column=0)尝试：Label(master,text="The&nbsp;file&nbsp;selected:&nbsp;"+file_path).grid(row=1,column=0)或者Label_1&nbsp;=&nbsp;Label(master,text="The&nbsp;file&nbsp;selected:&nbsp;"+file_path) Label_1.grid(row=1,column=0)也不要使用 Button = Button(master... 而是给变量一个唯一的名称

Answer 2

这是答案：from tkinter import *from tkinter import filedialogimport tkinter as tkimport pandas as pdimport pyodbcfrom sqlalchemy import create_engineimport urllibmaster = Tk()master.title("Demo GUI")master.geometry("900x400+150+150")master.resizable(0,0)def browse_file():&nbsp; &nbsp; global file_path&nbsp; &nbsp; global data_frame&nbsp; &nbsp; file_path = filedialog.askopenfilename(title = "Choose the file to upload")&nbsp; &nbsp; data_frame = pd.read_excel(file_path)Label = Label(master,text="Choose the file to upload").grid(row=0)Button = Button(master,text='Browse',command = browse_file).grid(row=0,column=1,pady=4)Label_1 = tk.Label(master,text="The file selected: "+file_path).grid(row=1,column=0)master.mainloop()

Answer 3

您所做的错误是一个错字：通过编写：Label&nbsp;=&nbsp;Label(master,text="Choose&nbsp;the&nbsp;file&nbsp;to&nbsp;upload").grid(row=0)您将grid调用结果分配给原始 tk.Label 类型 (&nbsp;Label)。网格回电是None因此，当您尝试创建 Label1 时，您实际上是在调用现在的 LabelNone只需将行替换为：&nbsp;l&nbsp;=&nbsp;Label(master,text="Choose&nbsp;the&nbsp;file&nbsp;to&nbsp;upload") &nbsp;l.grid(row=0)或者干脆Label(master,text="Choose&nbsp;the&nbsp;file&nbsp;to&nbsp;upload").grid(row=0)