计算器程序在 python 中返回 tkinter 错误答案

【问题标题】：Calculator program returns tkinter error in python计算器程序在 python 中返回 tkinter 错误
【发布时间】：2019-12-05 23:06:49
【问题描述】：

我正在尝试为计算器编写一个界面非常简单的程序。我仍然需要向它添加一些功能，但是，到目前为止，我测试了我的代码并且我得到了一个非常奇怪的 tkinter 错误。你能解释一下有什么问题吗？这是错误：

Exception in Tkinter callback
Traceback (most recent call last):
  File "/usr/lib/python3.6/tkinter/__init__.py", line 1705, in __call__
    return self.func(*args)
  File "/home/user/PycharmProjects/application /App1.py", line 8, in process
    number1 = Entry.get(E1)
  File "/usr/lib/python3.6/tkinter/__init__.py", line 2682, in get
    return self.tk.call(self._w, 'get')
AttributeError: 'NoneType' object has no attribute 'tk'

这是我的代码：

from tkinter import *
import tkinter
import tkinter.messagebox

def process():
    try :
        number1 = Entry.get(E1)
        number2 = Entry.get(E2)
        operator = Entry.get(E3)
        number1 = int (number1)
        number2 = int ( number2)
        if operator == "+" :
            answer = number1 + number2
        Entry.insert(E4,0,answer)
        print(answer)
    except ValueError:
        tkMessageBox.showwarning("Warning", "Please enter the value in integer")

top = tkinter.Tk()


L1  = Label(top, text ='My calculator',).grid(row = 0 , column = 1)
L2 = Label (top, text = "Number 1",).grid(row = 1 , column = 0)
L3 = Label (top, text = 'Number 2',).grid (row = 2, column = 0)
L4 = Label (top, text = 'Operator',) . grid (row=3, column = 0)
L5 = Label (top, text = 'Answer:',) . grid ( row = 4, column = 0)
E1 = Entry (top, bd = 5,).grid( row = 1 , column = 1)
# OR : E1.grid(row = 1, column = 1)
E2 = Entry ( top , bd = 5,).grid( row = 2, column = 1)
E3 = Entry ( top, bd = 5,).grid(row = 3, column = 1)
E4 = Entry ( top, bd = 5,).grid(row=4, column= 1 )
B = Button(top, text = "Submit here: ", command = process).grid (row = 5 , column = 1,)


top.mainloop()

【问题讨论】：

.grid() 方法返回None。你应该做的是E1= Entry(top, bd=5) 然后E1.grid(row=1,column=1) 让我知道这是否有帮助
@LazyCoder 是的，非常感谢。解释是 Entry.grid 只是在构建接口，而 Entry 也在返回一些东西？那么当我使用 entry.grid 时，你有什么想法吗？
@LazyCoder 好的，非常感谢。祝你有美好的一天！
Entry 创建一个对象。而 grid 调用一个返回 None 的方法。您需要 E1 的对象引用。就是这样。

标签： python tkinter calculator

【解决方案1】：

.grid() 方法返回None。你应该做的是E1= Entry(top, bd=5) 然后E1.grid(row=1,column=1) 错误将自行解决。

【讨论】：

【解决方案2】：

您确定Entry.get(...) 是正确的，即get 是静态/类方法，还是应该创建Entry 的实例并在其上调用get？

【讨论】：

【解决方案3】：

问题

.grid() 方法返回None，因此您将None 分配给您的变量。您应该将它们分成两行，例如：

代替：
```
E1 = Entry (top, bd = 5,).grid( row = 1 , column = 1)
```
使用：
```
E1 = Entry(top, bd = 5,)
E1.grid( row = 1 , column = 1)
```
使用from tkinter import messagebox 而不是import tkinter.messagebox。

最少的工作代码

下面应该是改动很小的工作代码

from tkinter import *
import tkinter
from tkinter import messagebox


def process():
    try :
        number1 = Entry.get(E1)
        number2 = Entry.get(E2)
        operator = Entry.get(E3)
        number1 = int(number1)
        number2 = int(number2)
        if operator == "+" :
            answer = number1 + number2
        Entry.insert(E4,0,answer)
        print(answer)
    except ValueError:
        messagebox.showwarning("Warning", "Please enter the value in integer")

top = tkinter.Tk()


L1 = Label(top, text ='My calculator',)
L1.grid(row = 0 , column = 1)

L2 = Label(top, text = "Number 1",)
L2.grid(row = 1 , column = 0)

L3 = Label(top, text = 'Number 2',)
L3.grid(row = 2, column = 0)

L4 = Label(top, text = 'Operator',)
L4.grid(row=3, column = 0)

L5 = Label(top, text = 'Answer:',)
L5.grid( row = 4, column = 0)

E1 = Entry(top, bd = 5,)
E1.grid( row = 1 , column = 1)

E2 = Entry( top , bd = 5,)
E2.grid( row = 2, column = 1)

E3 = Entry( top, bd = 5,)
E3.grid(row = 3, column = 1)

E4 = Entry( top, bd = 5,)
E4.grid(row=4, column= 1 )

B = Button(top, text = "Submit here: ", command = process)
B.grid(row = 5 , column = 1,)


top.mainloop()

【讨论】：