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python如何实现扑克牌交互式界面发牌程序

发布时间:2020-08-01 10:58:30 来源:亿速云 阅读:154 作者:小猪 栏目:开发技术

这篇文章主要讲解了python如何实现扑克牌交互式界面发牌程序,内容清晰明了,对此有兴趣的小伙伴可以学习一下,相信大家阅读完之后会有帮助。

注:图片自行在网上下载、替换即可

#coding=utf-8

class Card(): #扑克牌类
  points=['1','2','3','4','5','6','7','8','9','10','11','12','13']
  suits=['1','2','3','4'] #花色
  def __init__(self,points,suits):
    self.points=points
    self.suits=suits
  def __str__(self):
    rep=self.suits+'-'+self.points
    return rep
class Hand():
  def __init__(self):
    self.cards=[]
  def add(self,card):
    self.cards.append(card)
  def __str__(self):
    rep=''
    for card in self.cards:
      rep+=str(card)+'\t'
    return rep

class Poke(Hand):
  def generate_poke(self):
    for point in Card.points:
      for suit in Card.suits:
        self.add(Card(point,suit))
  def random_poke(self):
    import random
    random.shuffle(self.cards)
  def deal(self,hands,limit_hand=13):
    for rounds in range (limit_hand):
      for hand in hands:
        if self.cards:
          top=self.cards[0]
          self.cards.remove(top)
          hand.add(top)

def print_poke(players):
  results=[player.__str__() for player in players]
  pk=[]
  for result in results:
    pk1=result.rstrip()
    pk2=pk1.split('\t')
    pk.append(pk2)
  image_name=[]
  for i in range(len(pk)):
    for j in pk[i]:
      str_name='D:\pukepai\images\\{}.gif'.format(j)
      image_name.append(str_name)
  wj=[image_name[i:i+13] for i in range(len(image_name)) if i%13==0]
  return wj

def restart():
  tk.messagebox.showinfo("hello python","sdfg")

players=[Hand(),Hand(),Hand(),Hand()]
pockers=Poke()
pockers.generate_poke()
pockers.random_poke()
pockers.deal(players,13)
wj=print_poke(players) 

import tkinter as tk
win=tk.Tk()
win.title("扑克牌程序")
win.geometry('800x600')
cv=tk.Canvas(win,bg='red',width=800,height=600)
from PIL import Image,ImageTk
imgs=[]
(p1,p2,p3,p4)=([],[],[],[])
for i in range(4):
  for j in range(13):
    img=Image.open(wj[i][j])
    imgs.insert(i*13+j,ImageTk.PhotoImage(img))
p1=imgs[0:13]  
p2=imgs[13:26]
p3=imgs[26:39]
p4=imgs[39:52]
for x in range(0,13):
  cv.create_image((200+20*x,80),image=p1[x])
  cv.create_image((100,150+20*x),image=p2[x])
  cv.create_image((200+20*x,500),image=p3[x])
  cv.create_image((560,150+20*x),image=p4[x])

#添加重新发牌按钮
from tkinter.messagebox import *
bt1=tk.Button(win,text='重新发牌',width=60,height=40,command=restart)
bt1.place(x=290,y=380,width=60,height=40)

cv.pack()
win.mainloop()

看完上述内容,是不是对python如何实现扑克牌交互式界面发牌程序有进一步的了解,如果还想学习更多内容,欢迎关注亿速云行业资讯频道。

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