python - 通过单击按钮更改按钮文本不起作用

Question

我正在编写扫雷程序。我已经完成了所有的逻辑，现在我正在做 GUI。我正在使用 Tkinter。板上有这么多空间，我想自动创建所有这些按钮，我已经这样做了:

button_list = []

def create_buttons():
    # Create the buttons
    for x in range(code_squares):
        # Code_squares is how many squares are on the board
        new_button = Button(frame_list[x], text = "", relief = RAISED)
        new_button.pack(fill=BOTH, expand=1)
        new_button.bind("<Button-1>", lambda event: box_open(event, x))
        button_list.append(new_button)

def box_open(event, x):
    if box_list[x] == "M":
        # Checks if the block is a mine
        button_list[x].config(text="M", relief = SUNKEN)
        # Stops if it was a mine
        root.quit()
    else:
        # If not a mine, it changes the button text to the xth term in box_list, which is the amount of nearby mines.
        print("in")
        button_list[x].config(text=box_list[x], relief = SUNKEN)

打印语句只是一个测试。当我单击某个点时，它将执行打印语句，因此我知道它到达那里，但它不会更改按钮文本。非常感谢所有帮助!

Answer 1

我相信问题出在您尝试将 x 嵌入到 lambda 中的方式，请尝试一下，看看它是否可以解决您的问题:

from functools import partial

def create_buttons():
    for n in range(code_squares):
        # Code_squares is how many squares are on the board
        new_button = Button(frame_list[n], text="", relief=RAISED)
        new_button.pack(fill=BOTH, expand=1)
        new_button.bind("<Button-1>", partial(box_open, x=n))
        button_list.append(new_button)

def box_open(event, x):
    if box_list[x] == "M":
        # Checks if the block is a mine
        button_list[x].config(text="M", relief=SUNKEN)
        # Stops if it was a mine
        root.quit()
    else:
        # If not a mine, it changes the button text to the xth
        # term in box_list, which is the amount of nearby mines.
        button_list[x].config(text=box_list[x], relief=SUNKEN)

button_list = []