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原标题:Generate Tkinter Buttons dynamically

我想产生<代码>ns>,tkinter Buttons,其中做的是不同的事。 我有这一法典:

import Tkinter as tk

for i in range(boardWidth):
    newButton = tk.Button(root, text=str(i+1),
                    command=lambda: Board.playColumn(i+1, Board.getCurrentPlayer()))
    Board.boardButtons.append(newButton)

如果<条码>登上5>,如果被点击所有<条码>/条码>,则请上表5。

I need the first button to do Board.playColumn (1, Board.getCurrentPlayer(), the second to do Board.playColumn(2, Board.getCurrentPlayer().

最佳回答

我认为,问题在于,lambda>正在取到i之后的i<>。 这应当确定(未测试):

import Tkinter as tk

for i in range(boardWidth):
    newButton = tk.Button(root, text=str(i+1),
                    command=lambda j=i+1: Board.playColumn(j, Board.getCurrentPlayer()))
    Board.boardButtons.append(newButton)

<>Update>

BTW在这项工作中添加了一个论点,在“条码”栏目/代码”功能中添加一个词句,其设定值为“条码>ii的最后价值,而在其内部的表述后来执行时,则通过封闭方式。

问题回答

您的问题是,你以同一名称空间制造了lambda 的物体,而lambda则指外层空间的名称。 这意味着它们不会变成封闭,直到以后它们不会储存对物体的提及...... 发生这种情况时,所有拉姆布达斯都将提及<代码>i的最后价值。

3. 利用电击厂确定:

import Tkinter as tk

def callbackFactory(b, n):
    def _callback():
        return b.playColumn(n, b.getCurrentPlayer())
    return _callback

for i in range(boardWidth):
    newButton = tk.Button(root, text=str(i+1), 
        command=callbackFactory(Board, i+1))
    Board.boardButtons.append(newButton)

另一种想法是,将i作为缺省理由的现值储存在lambda的标的上,而不是依靠封闭行为储存提及:

for i in range(boardWidth):
    newButton = tk.Button(root, text=str(i+1), 
        command=lambda x=i: Board.playColumn(x+1, Board.getCurrentPlayer()))
    Board.boardButtons.append(newButton)




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