例如，當我按下“100”時，在文本框中輸出“001”。我嘗試在索引中使用 -1 但同樣的事情仍然發生，我也嘗試執行 .insert(0.end, num) 但它會引發錯誤。我怎樣才能讓數字總是在輸出的末尾輸入。此外，這是使用 tkinter 輸出數字的最佳方式還是有其他方式？from tkinter import *import operatorwindow = Tk()window.title('Calculator')def click(num): output.insert(0.0, num) #numbers not properly inputted (bug)#output for calculatoroutput = 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 operatorsb_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))#numbersb_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))
2 回答
慕哥9229398
TA貢獻1877條經驗 獲得超6個贊
索引“end”表示小部件中最后一個字符之后的位置。
output.insert("end", num)來自官方文檔：
end - 表示條目字符串中最后一個字符之后的字符。這相當于指定一個等于條目字符串長度的數字索引。
子衿沉夜
TA貢獻1828條經驗 獲得超3個贊
你工作太辛苦了。從圖形的角度來看，您必須考慮到您的按鈕都是完全相同的東西。您只需幾行就可以構建整個界面。由于每個按鈕都會調用calc，只需編寫條件語句來calc處理各種可能性。您可以在該函數中構建計算器的全部功能。
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()
我根據剛才給你的例子制作了一個完全可用的 OOP 版本的計算器。它可能并不完美。我只花了 10 分鐘。我添加了鍵綁定，因此您不必單擊按鈕。你可以擴展它，從中學習，忽略它......任何讓你開心的事情。
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()
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