我知道這個問題很多,但我似乎無法讓我的代碼工作。python TKinter'int'/'str'對象沒有屬性'追加'
作爲一個預測,我試圖建立一個簡單的計算器。但我有點卡住了。這是我的代碼。
import Tkinter as tk
import tkMessageBox
top = tk.Tk()
def helloCallBack(x):
counter = 0
counter.append(x)
tkMessageBox.showinfo("result", counter)
one = tk.Button (top, text = "1", command = lambda: helloCallBack(1))
two = tk.Button (top, text = "2", command = lambda: helloCallBack(2))
three = tk.Button (top, text = "3", command = lambda: helloCallBack(3))
four = tk.Button (top, text = "4", command = lambda: helloCallBack(4))
five = tk.Button (top, text = "5", command = lambda: helloCallBack(5))
six = tk.Button (top, text = "6", command = lambda: helloCallBack(6))
seven = tk.Button (top, text = "7", command = lambda: helloCallBack(7))
eight = tk.Button (top, text = "8", command = lambda: helloCallBack(8))
nine = tk.Button (top, text = "9", command = lambda: helloCallBack(9))
zero = tk.Button (top, text = "9", command = lambda: helloCallBack(0))
one.pack()
two.pack()
three.pack()
four.pack()
five.pack()
six.pack()
seven.pack()
eight.pack()
nine.pack()
zero.pack()
top.mainloop()
我目前得到「詮釋」對象有沒有屬性「追加」
這意味着你不能使用數字附加命令?
如果是的話如何才能做到這一點,如果我按下其中一個按鈕,它將該數字添加到櫃檯,所以如果你按下按鈕一,二,五,你會得到0125我也試過這樣做與
counter = ""
,但只是給了同樣的錯誤,但以「海峽」對象有沒有屬性「追加」
我是新來的Python和任何幫助將不勝感激。
'append'函數是列表。嘗試'counter + = x' – bunji
Mb嘗試使用'counter = 0 counter + = str(x)'而不是.append。它應該工作,因爲我們預先定義了它的類型和連接字符串,而不是整數。 – Grynets