如何正确地将我在计算器中按下的数字插入到最后一个索引处的文本小部件中？

Question

When I pres "100" for example, in the text box it outputs "001". I try using -1 in the index but the same thing still happens, and I tried doing.insert(0. end, num) as well but it throws an error. How can I have the numbers always input at the end of the output. Also, is this the best way to output numbers with tkinter or are there other ways if any?

from tkinter import *
import operator

window = Tk()
window.title('Calculator')

def click(num):
    output.insert(0.0, num) #numbers not properly inputted (bug)

#output for calculator
output = Text(window, font = 'none 12 bold', height = 4, width = 25, wrap = 'word')
output.grid(row = 0, column = 0, columnspan = 4, pady = 10)

###buttons
#clear and operators
b_clear = Button(window, text = 'C', width = 7, height = 3)
b_clear.grid(row = 1, column = 2, padx = (10, 0))

b_div = Button(window, text = '/', width = 7, height = 3)
b_div.grid(row = 1, column = 3, padx = 10)

b_mult = Button(window, text = '*', width = 7, height = 3)
b_mult.grid(row = 2, column = 3)

b_subt = Button(window, text = '-', width = 7, height = 3)
b_subt.grid(row = 3, column = 3)

b_add = Button(window, text = '+', width = 7, height = 3)
b_add.grid(row = 4, column = 3)

b_equal = Button(window, text = '=', width = 7, height = 3)
b_equal.grid(row = 5, column = 3, pady = (0, 10))

#numbers
b_9 = Button(window, text = '9', width = 7, height = 3, command = lambda: click(9))
b_9.grid(row = 2, column = 2, padx = (10, 0), pady = 10)

b_8 = Button(window, text = '8', width = 7, height = 3, command = lambda: click(8))
b_8.grid(row = 2, column = 1)

b_7 = Button(window, text = '7', width = 7, height = 3, command = lambda: click(7))
b_7.grid(row = 2, column = 0, padx = 10)

b_6 = Button(window, text = '6', width = 7, height = 3, command = lambda: click(6))
b_6.grid(row = 3, column = 2, padx = (10, 0))

b_5 = Button(window, text = '5', width = 7, height = 3, command = lambda: click(5))
b_5.grid(row = 3, column = 1)

b_4 = Button(window, text = '4', width = 7, height = 3, command = lambda: click(4))
b_4.grid(row = 3, column = 0)

b_3 = Button(window, text = '3', width = 7, height = 3, command = lambda: click(3))
b_3.grid(row = 4, column = 2, padx = (10, 0), pady = 10)

b_2 = Button(window, text = '2', width = 7, height = 3, command = lambda: click(2))
b_2.grid(row = 4, column = 1)

b_1 = Button(window, text = '1', width = 7, height = 3, command = lambda: click(1))
b_1.grid(row = 4, column = 0)

b_0 = Button(window, text = '0', width = 7, height = 3, command = lambda: click(0))
b_0.grid(row = 5, column = 0, pady = (0, 10))

b_decimal = Button(window, text = '.', width = 7, height = 3)
b_decimal.grid(row = 5, column = 1, pady = (0, 10))

b_negative = Button(window, text = '-', width = 7, height = 3)
b_negative.grid(row = 5, column = 2, padx = (10, 0), pady = (0, 10))

#run calculator
window.mainloop()

Answer 1

The index "end" represents the position just after the last character in the widget.

output.insert("end", num) 

From the official documentaton:

end - Indicates the character just after the last one in the entry's string. This is equivalent to specifying a numerical index equal to the length of the entry's string.

Answer 2

You are working way too hard. You have to consider that your buttons are all the exact same thing, from a graphical standpoint. You could build your entire interface in just a few lines. Since every button is going to call calc , simply write conditional statements in calc to handle the various possibilities. You could build the entire functionality of a calculator in that one function.

import tkinter as tk 

window = tk.Tk()
window.title('Calculator')

#output for calculator
output = tk.Text(window, font = 'none 12 bold', height=4, width=25, wrap='word')
output.grid(row=0, column=0, columnspan=4, pady=10)

def calc(data):
    if data.isnumeric() or data == '.':
        output.insert('end', data)
    elif data in ['-', '+', '*', '/']:
        #write code for handling operators
        pass #delete this line
    elif data == '=':
        #write code for handling equals
        pass #delete this line
    elif data == 'pos':
        if output.get('1.0', '1.1') == '-':
            output.delete('1.0', '1.1') 
    elif data == 'neg':
        if output.get('1.0', '1.1') != '-':
            output.insert('1.0', '-')
    elif data in 'CE':
        if 'C' in data:
            output.delete('1.0', 'end')
        if 'E' in data:
            #clear your storage
            pass #delete this line

btn = dict(width=7, height=3)
pad = dict(padx=5, pady=5)

#all of your buttons
for i, t in enumerate(['pos', 'neg', 'C', 'CE', '7', '8', '9', '/', '4', '5', '6', '*', '1', '2', '3', '-', '0', '.', '+', '=']):
    tk.Button(window, text=t, command=lambda d=t: calc(d), **btn).grid(row=i//4+1, column=i%4, **pad)

#run calculator
window.mainloop()

---

I made a fully working OOP version of your calculator based on the example I just gave you. It's probably not perfect. I only spent 10 minutes on it. I added key bindings so you don't have to click buttons. You can expand this, learn from it, ignore it... whatever makes you happy.

import tkinter as tk

class App(tk.Tk):
    def __init__(self):
        tk.Tk.__init__(self)
        self.oper   = ['/', '*', '-', '+']
        self.queue  = []
        self.result = 0
        self.equate = False
        
        self.output = tk.Text(self, font='Consolas 18 bold', height=2, width=20, wrap='word')
        self.output.grid(row=0, column=0, columnspan=4, pady=8, padx=4)
        
        btn = dict(width=6, height=3, font='Consolas 12 bold')
        pad = dict(pady=2)
        
        special = dict(neg='<Shift-Key-->',pos='<Shift-Key-+>',C='<Key-Delete>',CE='<Key-End>')
        
        for i, t in enumerate(['pos', 'neg', 'C', 'CE', '7', '8', '9', '/', '4', '5', '6', '*', '1', '2', '3', '-', '0', '.', '+', '=']):
            tk.Button(self, text=t, command=lambda d=t: self.calc(d), **btn).grid(row=i//4+1, column=i%4, **pad)
            if t.isnumeric() or t in self.oper or t == '.':
                self.bind_all(f'<Key-{t}>', lambda e, d=t: self.calc(d))
            elif t == '=':
                self.bind_all('<Return>', lambda e, d=t: self.calc(d))
            else:
                self.bind_all(special[t], lambda e, d=t: self.calc(d))
                
    def calc(self, input):
        print(input)
        if input.isnumeric() or input == '.':
            self.output.insert('end', input)
        elif input == 'pos':
            if self.output.get('1.0', '1.1') == '-':
               self.output.delete('1.0', '1.1') 
        elif input == 'neg':
            if self.output.get('1.0', '1.1') != '-':
               self.output.insert('1.0', '-')
        elif input in self.oper and (t := self.output.get('1.0', 'end-1c')):
            if not self.equate:
                self.queue.append(t)
            self.queue.append(input)
            self.output.delete('1.0', 'end')
            self.equate = False
        elif input == '=' and len(self.queue):
            self.equate = True
            if self.queue[-1] in self.oper:
                self.queue.append(self.output.get('1.0', 'end-1c'))
            elif len(self.queue) > 2:
                self.queue = self.queue+self.queue[-2:]
                    
            self.result = str(eval(' '.join(self.queue)))
            self.output.delete('1.0', 'end')
            self.output.insert('end', self.result)
        elif input in 'CE':
            if 'C' in input:
                self.output.delete('1.0', 'end')
            if 'E' in input:
                self.queue = []
            
                    
if __name__ == '__main__':
    app = App()
    app.title('Calcsturbator')
    app.resizable(width=False, height=False)
    app.mainloop()