Python Tkinter Entry()计算

2024-06-01 13:38:26 发布

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我目前正试图在tkinter的帮助下使用python构建一个GUI计算器。我已经成功地设置了我所有的按钮,当我按下按钮时,它们会与我的entry()栏交互(例如:按按钮5,5会出现在entry()中)。在

剩下要做的就是实际执行entry()中出现的数学方程。例如,如果我在输入栏中输入5+5*2,那么在更新后,如何使答案出现在entry()中?所以基本上我只要求有人帮我实现这个目标!在

我提供了下面的代码和计算器的屏幕截图。另外,请不要将此问题报告为重复,也不要将此问题标记为帖子(我重新提出此问题是因为我上一个问题没有帮助,所以请帮助一个年轻的python程序员!)。同时我要求你不要给我提供链接,因为我已经阅读了tkinter上所有可能的链接,但仍然不知道我在做什么。在

这是我的代码:

import sys
from tkinter import *
from PIL import Image, ImageTk

#ACTIONS:
def clear():
    #Action: Clears the entry().
    txtDisplay.delete(0,END);
    return;

def update_Entry(inputValue):
    #Updates the entry when a button is pressed.
    currentText = num1.get();
    update = num1.set(currentText + inputValue);

#Parent Window.
root = Tk();
root.title('Calculator ++ [1.7.2]');
root.geometry('350x450');

#Main entry.
num1 = StringVar();
txtDisplay = Entry(root, textvariable = num1, relief=RIDGE, bd = 10, width=33, insertwidth = 1, font = 40, justify=RIGHT);
txtDisplay.place(x=15, y=10);
txtDisplay.focus();

print(txtDisplay.get())

#Buttons:
zeroButton = Button(root, text='0', width=20, height=3, bg='LightBlue', fg='red', command=lambda:update_Entry('0'));
zeroButton.place(x=17,y=382);
oneButton = Button(root, text='1', width=8, height=3, bg='LightBlue', fg='red', command=lambda:update_Entry('1'));
oneButton.place(x=17, y=302);
twoButton = Button(root, text='2', width=8, height=3, bg='LightBlue', fg='red', command=lambda:update_Entry('2'));
twoButton.place(x=100, y=302);
threeButton = Button(root, text='3', width=8, height=3, bg='LightBlue', fg='red', command=lambda:update_Entry('3'));
threeButton.place(x=182, y=302);
fourButton = Button(root, text='4', width=8, height=3, bg='LightBlue', fg='red', command=lambda:update_Entry('4'));
fourButton.place(x=17, y=222);
fiveButton = Button(root, text='5', width=8, height=3, bg='LightBlue', fg='red', command=lambda:update_Entry('5'));
fiveButton.place(x=100, y=222);
sixButton = Button(root, text='6', width=8, height=3, bg='LightBlue', fg='red', command=lambda:update_Entry('6'));
sixButton.place(x=182, y=222);
sevenButton = Button(root, text='7', width=8, height=3, bg='LightBlue', fg='red', command=lambda:update_Entry('7'));
sevenButton.place(x=17, y=142);
eightButton = Button(root, text='8', width=8, height=3, bg='LightBlue', fg='red', command=lambda:update_Entry('8'));
eightButton.place(x=100, y=142);
ninthButton = Button(root, text='9', width=8, height=3, bg='LightBlue', fg='red', command=lambda:update_Entry('9'));
ninthButton.place(x=182, y=142);

decimalButton = Button(root, text='.', width=8, height=3, bg='powder blue', command=lambda:update_Entry('.'));
decimalButton.place(x=182, y=382);
equalButton = Button(root, text='=', width=8, height=8, bg='Lightgreen', command=lambda:update_Entry('='));
equalButton.place(x=264, y=307);
plusButton = Button(root, text='+', width=8, height=3, bg='gray', command=lambda:update_Entry('+'));
plusButton.place(x=264, y=222);
minusButton = Button(root, text='-', width=8, height=3, bg='gray', command=lambda:update_Entry('-'));
minusButton.place(x=264, y=142);
multiplyButton = Button(root, text='x', width=8, height=3, bg='gray', command=lambda:update_Entry('*'));
multiplyButton.place(x=264, y=66);
divideButton = Button(root, text='÷', width=8, height=3, bg='gray', command=lambda:update_Entry('/'));
divideButton.place(x=182, y=66);
clearButton = Button(root, text='Clear (CE)', width=20, height=3, command = clear, bg='Orange');
clearButton.place(x=17, y=66);

#Locks the parent windows size.
root.maxsize(350,450);
root.minsize(350,450);

#Parent window's background color:
root.configure(background = 'black');
root.mainloop();

截图:

Screenshot


Tags: lambdatextupdateplacebuttonredrootwidth
1条回答
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1楼 · 发布于 2024-06-01 13:38:26

对于这样简单的事情,请尝试将其作为您的equalButtoncommand

def evaluate():
    currentText = num1.get()
    try:
        num1.set(str(eval(currentText)))
    except SyntaxError:
        num1.set('<ERROR>')

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