我遇到了 tkinter def 无法启动的问题

Question

当我运行并按下开始按钮时我制作了 facebook 脚本我收到了这个错误我用 Tkinter 制作了这个 GUI 界面我正在使用 python 3 我的错误是当我点击开始按钮时他没有启动谷歌浏览器当我禁用 Tkinter 并通过命令行工作时，他可以正常工作，您可以看到错误屏幕截图：

import pyautogui as pg
import time
from selenium import webdriver
from getpass import getpass
from tkinter import *

window = Tk()
window.title('Facebook Sharing')
text = Label(window, text='Facebook Sharing Script', font="Source_Sans_Pro 20 bold")
text.pack()
frame = Frame(window)
frame.pack()

one = Frame(window)
one.pack( side = TOP )

frontframe = Frame(window)
frontframe.pack( side = TOP )


bottomframe = Frame(window)
bottomframe.pack( side = TOP )

lastframe = Frame(window)
lastframe.pack( side = TOP )

last = Frame(window)
last.pack( side = TOP )

Lable1 = Label(frame, text="Enter Your UserName!    ", font='Source_Sans_Pro 11')
Lable1.pack( side = LEFT)
user_name = Entry(frame, bd =5)
user_name.pack(side = RIGHT)

Lable2 = Label(frontframe, text="Enter Your Passowrd!     ", font='Source_Sans_Pro 11')
Lable2.pack( side = LEFT)
Password = Entry(frontframe, bd =5)
Password.pack(side = RIGHT)

Lable3 = Label(bottomframe, text="Enter Your Description!   ", font='Source_Sans_Pro 11')
Lable3.pack( side = LEFT)
description = Entry(bottomframe, bd =5)
description.pack(side = RIGHT)

Lable4 = Label(lastframe, text="Enter Your Keyword!       ", font='Source_Sans_Pro 11')
Lable4.pack( side = LEFT)
keyword = Entry(lastframe, bd =5)
keyword.pack(side = RIGHT)

Lable5 = Label(one, text="Enter Your Post Url!         ", font='Source_Sans_Pro 11')
Lable5.pack( side = LEFT)
post_url = Entry(one, bd =5)
post_url.pack(side = RIGHT)

Button_Start = Button(last, text="Start", fg="black", width=50, command=all)
Button_Start.pack( side = BOTTOM)


pg.FAILSAFE = True

#this is hidden def
def all():

print("Done")
window.mainloop()

错误！

Exception in Tkinter callback
Traceback (most recent call last):
  File "C:\User`enter code here`s\Hamza Lachi\AppData\Local\Programs\Python\Python37-32\lib\tkinter\__init__.py", line 1705, in __call__
    return self.func(*args)
TypeError: all() takes exactly one argument (0 given)

Answer 1

所以All() 是原生 Python 函数，你可以在下面的链接中阅读更多关于它的链接 all(iterable)，这就是问题所在，Python 正在调用他的原生函数而不是你创建的函数。

您可以通过更改函数名称来解决此问题。

了解更多信息：

• Built-in Functions
• List of Keywords in Python