python实现带界面的井字棋

1,268 阅读2分钟

image.png

今天我们用python+tkinter安装带界面的井字棋,效果如上图所示。

Tkinter 是 Python 的标准 GUI 库。Python 使用 Tkinter 可以快速的创建 GUI 应用程序。由于 Tkinter 是内置到 python 的安装包中、只要安装好 Python 之后就能 import Tkinter 库、而且 IDLE 也是用 Tkinter 编写而成、对于简单的图形界面 Tkinter 还是能应付自如。

pip install tkinter

首先安装tkinter

root= Tk()
root.title('井字棋')

digits = [1,2,3,4,5,6,7,8,9]
mark = ''count = 0
panels = ["panel"]*10

初始化窗口

Label(root,text="player1 : X",font="times 15").grid(row=0,column=1)
Label(root,text="player2 : O",font="times 15").grid(row=0,column=2)

button1=Button(root,width=15,font=('Times 16 bold'),height=7,command=lambda:checker(1))
button1.grid(row=1,column=1)
button2=Button(root,width=15,height=7,font=('Times 16 bold'),command=lambda:checker(2))
button2.grid(row=1,column=2)

button3=Button(root,width=15,height=7,font=('Times 16 bold'),command=lambda: checker(3))
button3.grid(row=1,column=3)
button4=Button(root,width=15,height=7,font=('Times 16 bold'),command=lambda: checker(4))
button4.grid(row=2,column=1)

button5=Button(root,width=15,height=7,font=('Times 16 bold'),command=lambda: checker(5))
button5.grid(row=2,column=2)
button6=Button(root,width=15,height=7,font=('Times 16 bold'),command=lambda: checker(6))
button6.grid(row=2,column=3)

button7=Button(root,width=15,height=7,font=('Times 16 bold'),command=lambda: checker(7))
button7.grid(row=3,column=1)
button8=Button(root,width=15,height=7,font=('Times 16 bold'),command=lambda: checker(8))
button8.grid(row=3,column=2)

button9=Button(root,width=15,height=7,font=('Times 16 bold'),command=lambda: checker(9))
button9.grid(row=3,column=3)


root.mainloop()

定义按钮,很明显上面的井字棋就是9个按钮。这里定义9个按钮。

def win(panels,sign):
 return ((panels[1] == panels[2] == panels [3] == sign)
   or (panels[1] == panels[4] == panels [7] == sign)
   or (panels[1] == panels[5] == panels [9] == sign)
   or (panels[2] == panels[5] == panels [8] == sign)
   or (panels[3] == panels[6] == panels [9] == sign)
   or (panels[3] == panels[5] == panels [7] == sign)
   or (panels[4] == panels[5] == panels [6] == sign) 
   or (panels[7] == panels[8] == panels [9] == sign))

检查获胜的条件,上面其中一种情况都是获胜的。

def checker(digit):
 global count, mark, digits
 if digit==1 and digit in digits:
  digits.remove(digit)
  if count%2==0:
   mark ='X'
   panels[digit]=mark
  elif count%2!=0:
   mark = 'O'
   panels[digit]=mar
  button1.config(text = mark)
  count = count+1
  sign = mark
  if(win(panels,sign) and sign=='X'):
   msg.showinfo("Result","Player1 wins")
   root.destroy()
  elif(win(panels,sign) and sign=='O'):
   msg.showinfo("Result","Player2 wins")
   root.destroy()

最后这里是点击下棋,然后用win函数判断是否结束比赛赢得棋局。

欢迎和我讨论有关程序的问题,也可以答疑。关注公众号:诗一样的代码,交一个朋友。