井字游戏 Python Tkinter 显示“X”和“Y”

Question

我已经开始用 Python 为 Tkinter 制作一个井字游戏。基本上在这一点上，它并不需要获胜，而只是具有基本功能。左键单击放置一个“X”，右键单击放置一个“O”。按“C”应该清除电路板，并且应该使用标签和绑定来完成。我在显示“X”和“O”时遇到了问题，我确实画了板。我尝试单击第一行，但没有任何反应。

它应该像这样工作：https://youtu.be/0x6HKuyXAuU

from tkinter import *
from tkinter import ttk


root = Tk()
root.title("Tic Tac Toe")

message = StringVar()

# This is the portion that draws out the grid used in Tic Tac Toe

label1 = ttk.Label(root, text="")
label1.grid(row=1, column=0)
label2 = ttk.Label(root, text="|")
label2.grid(row=1, column=1)
label3 = ttk.Label(root, text="")
label3.grid(row=1, column=2)
label4 = ttk.Label(root, text="|")
label4.grid(row=1, column=3)
label5 = ttk.Label(root, text="-")
label5.grid(row=4, column=1)
label6 = ttk.Label(root, text="-")
label6.grid(row=4, column=2)
label7 = ttk.Label(root, text="-")
label7.grid(row=4, column=3)
label8 = ttk.Label(root, text="-")
label8.grid(row=4, column=4)
label9 = ttk.Label(root, text="       ")
label9.grid(row=5, column=0)
label10 = ttk.Label(root, text="|")
label10.grid(row=5, column=1)
label11 = ttk.Label(root, text="       ")
label11.grid(row=5, column=2)
label12 = ttk.Label(root, text="|")
label12.grid(row=5, column=3)
label13 = ttk.Label(root, text="-")
label13.grid(row=6, column=1)
label14 = ttk.Label(root, text="-")
label14.grid(row=6, column=2)
label15 = ttk.Label(root, text="-")
label15.grid(row=6, column=3)
label16 = ttk.Label(root, text="-")
label16.grid(row=6, column=4)
label17 = ttk.Label(root, text="       ")
label17.grid(row=7, column=0)
label18 = ttk.Label(root, text="|")
label18.grid(row=7, column=1)
label19 = ttk.Label(root, text="       ")
label19.grid(row=7, column=2)
label20 = ttk.Label(root, text="|")
label20.grid(row=7, column=3)

#This is the code that draws out the O and X in this the first row
label21 = ttk.Label(root, text="")
label21.grid(row=1, column=0)
label21.bind("<Button-1>", lambda e: message.set("X"))
label21.bind("<Button-3>", lambda e: message.set("O"))
label22 = ttk.Label(root, text="")
label22.grid(row=1, column=1)
label22.bind("<Button-1>", lambda e: message.set("X"))
label22.bind("<Button-3>", lambda e: message.set("O"))
label22.grid(row=1, column=3)
label22.bind("<Button-1>", lambda e: message.set("X"))
label22.bind("<Button-3>", lambda e: message.set("O"))

root.mainloop()

Answer 1

使用循环而不是像这样创建大量变量和标签：

这是一个你可以改进的例子：

from tkinter import *

def updateX(event):
    event.widget['text'] = 'X'

def updateO(event):
    event.widget['text'] = 'O'
    
root = Tk()
root.title("Tic Tac Toe")

for x in range(5):
    for y in range(5):
        label = Label(root)

        if x%2==0:
            if y%2!= 0:
                label['text'] = '|'

            else:
    
                label.bind('<1>', updateX)
                label.bind('<Button-3>', updateO)
                root.grid_columnconfigure(y, weight=1)
                root.grid_rowconfigure(x, weight=1)

        else:
            label['text']='-'

        label.grid(column=y, row=x, sticky="nsew")

root.mainloop()

你的 clear 函数可以是这样的：

def clear(event):
    for item in root.winfo_children():
        if isinstance(item, Label) and item['text'] in ['O', 'X']:
            item['text'] = ''

别忘了将<c>绑定到root。

Answer 2

试试这个：

from functools import partial
import tkinter as tk


def change_text(button, text, event=None):
    # Only if the button is empty
    if button.cget("text") == "":
        # Change the button's text to whatever is in the variable `text`
        button.config(text=text)

def clear(event=None):
    for button in buttons:
        button.config(text="")


# Create the tkinter window
root = tk.Tk()
# A list to store all of the buttons
buttons = []

for i in range(9):
    # Create the button:
    button = tk.Label(root, text="", width=3, font=("", 15), bd=3,
                      relief="groove")

    # 1st parameter of `partial` is the function. The rest are just arguments
    command = partial(change_text, button, "X")
    button.bind("<Button-1>", command) # Left click

    command = partial(change_text, button, "O")
    button.bind("<Button-3>", command) # Right click

    # Display the button:
    button.grid(row=i//3, column=i%3)
    # Add the button to the list of buttons:
    buttons.append(button)

clear_button = tk.Label(root, text="Clear screen", width=3, font=("", 15), bd=3,
                        relief="groove")
clear_button.grid(row=4, column=0, columnspan=3, sticky="news")
clear_button.bind("<Button-1>", clear) # Left click

# Run tkinter's mainloop
root.mainloop()

有关partial 工作原理的更多信息，请阅读this。

Answer 3

您的绑定lambda 只是更新了那些标签从未使用过的message 变量。

您可以使用带有边框的Label 小部件来模拟游戏板。最好使用list 或dict 来存储这些标签以供以后参考，例如通过按c 字符来清除游戏板。

下面是一个修改过的例子：

import tkinter as tk
from tkinter import ttk

root = tk.Tk()
root.title("Tic Tac Toe")

def set_cell(lbl, mark):
    if lbl["text"] == "":
        lbl["text"] = mark

def clear(event):
    for lbl in cells.values():
        lbl["text"] = ""

cells = {} # for storing the labels
# create the 3 x 3 board using labels
for row in range(3):
    for col in range(3):
        lbl = ttk.Label(root, width=3, borderwidth=1, relief="solid", font="Arial 32 bold", anchor="c")
        lbl.grid(row=row, column=col, sticky="nsew")
        lbl.bind("<Button-1>", lambda e: set_cell(e.widget, "X"))
        lbl.bind("<Button-3>", lambda e: set_cell(e.widget, "O"))
        cells[row,col] = lbl # store the label

root.bind("c", clear)
root.mainloop()