在TKINTER CALCULATOR中添加CLEAR ENTRY按钮

问题描述

我正在用Tkinter（Python）编写一个计算器。我要添加一个“清除条目”按钮，以清除用户给出的“条目”。不要对“全部清除”按钮感到困惑。

示例： 用户输入：-35 + 87 = 62 并按CE按钮，因此应该显示：-35 + 87，如果再次按该按钮，则应该得到35，依此类推，直到0。

我尝试了很多，最后，我来找你。 这是基本代码，（它没有任何CE按钮，您可以添加一个）。您可以使用正则表达式。

from tkinter import *
expression = "" 
def press(num): 
    global expression 
    expression = expression + str(num) 
    equation.set(expression) 
  
  

def equalpress(): 
    try: 
        global expression 
        total = str(eval(expression)) 
        equation.set(total) 
        expression = "" 
    except: 
        equation.set(" error ") 
        expression = "" 
def clear(): 
    global expression 
    expression = "" 
    equation.set("") 
  

if __name__ == "__main__": 
    # create a GUI window 
    gui = Tk() 
  
    # set the background colour of GUI window 
    gui.configure(background="light green") 
  
    # set the title of GUI window 
    gui.title("Simple Calculator") 
  
    # set the configuration of GUI window 
    gui.geometry("270x150") 
  
    # StringVar() is the variable class 
    # we create an instance of this class 
    equation = StringVar() 
  
    # create the text entry box for 
    # showing the expression . 
    expression_field = Entry(gui,textvariable=equation) 
  
    expression_field.grid(columnspan=4,ipadx=70) 
  
    equation.set('enter your expression') 
  
    button1 = Button(gui,text=' 1 ',fg='black',bg='red',command=lambda: press(1),height=1,width=7) 
    button1.grid(row=2,column=0) 
  
    button2 = Button(gui,text=' 2 ',command=lambda: press(2),width=7) 
    button2.grid(row=2,column=1) 
  
    button3 = Button(gui,text=' 3 ',command=lambda: press(3),width=7) 
    button3.grid(row=2,column=2) 
  
    button4 = Button(gui,text=' 4 ',command=lambda: press(4),width=7) 
    button4.grid(row=3,column=0) 
  
    button5 = Button(gui,text=' 5 ',command=lambda: press(5),width=7) 
    button5.grid(row=3,column=1) 
  
    button6 = Button(gui,text=' 6 ',command=lambda: press(6),width=7) 
    button6.grid(row=3,column=2) 
  
    button7 = Button(gui,text=' 7 ',command=lambda: press(7),width=7) 
    button7.grid(row=4,column=0) 
  
    button8 = Button(gui,text=' 8 ',command=lambda: press(8),width=7) 
    button8.grid(row=4,column=1) 
  
    button9 = Button(gui,text=' 9 ',command=lambda: press(9),width=7) 
    button9.grid(row=4,column=2) 
  
    button0 = Button(gui,text=' 0 ',command=lambda: press(0),width=7) 
    button0.grid(row=5,column=0) 
  
    plus = Button(gui,text=' + ',command=lambda: press("+"),width=7) 
    plus.grid(row=2,column=3) 
  
    minus = Button(gui,text=' - ',command=lambda: press("-"),width=7) 
    minus.grid(row=3,column=3) 
  
    multiply = Button(gui,text=' * ',command=lambda: press("*"),width=7) 
    multiply.grid(row=4,column=3) 
  
    divide = Button(gui,text=' / ',command=lambda: press("/"),width=7) 
    divide.grid(row=5,column=3) 
  
    equal = Button(gui,text=' = ',command=equalpress,width=7) 
    equal.grid(row=5,column=2) 
  
    clear = Button(gui,text='Clear',command=clear,width=7) 
    clear.grid(row=5,column='1') 
  
    Decimal= Button(gui,text='.',command=lambda: press('.'),width=7) 
    Decimal.grid(row=6,column=0) 
    gui.mainloop() 

解决方法

在“等式”字段中，仅允许使用数字，运算符和小数点。要清除最新的条目，请从方程式的末尾搜索第一个运算符（+-* /），然后截断字符串。

将此功能用于ClearEntry按钮：

def clearentry(): 
    global expression 
    for i,c in enumerate(expression[::-1]): # reverse string
       if c in ['+','-','/','*']:  # find first operator 
           expression = expression[:-i-1]  # truncate string
           break
    else:
      expression = "" # no operator found
    equation.set(expression) 

,

我将此作为答案发布，因为由于我的声誉低，我无法发表任何评论。所以，我只是对 Mike67 的回答中遗漏的一点做了一点补充。

如果您在对等式求和后按清除输入按钮，则会显示等式。例如，如果您对 2+3 进行求和，并且屏幕上显示 5。这将显示为 2+3。我只是在这两个函数中添加了几行：

def equalpress():
    global backup_expression
    try:
        global expression
        total = str(eval(expression))
        equation.set(total)
        backup_expression = expression
        expression = ""

    except:
        equation.set(" error ")
        expression = ""

def clearentry():
    global expression,backup_expression
    for i,'*']:  # find first operator
           expression = expression[:-i-1]  # truncate string
           break
    else: # no operator found
        expression = backup_expression
    equation.set(expression)