Python多人游戏不存在和交叉

Python多人游戏不存在和交叉,第1张

所以我在不久前在学校做了一个单独的玩家,并且用Python进行了交叉。

score=[[' ',' ',' '],[' ',' ',' '],[' ',' ',' ']]
global attempts
attempts = 0
from random import randint
from time import sleep

def grid(): #draws playing grid
    hideturtle()
    speed(0)
    pensize(0)
    penup()
    setpos(-200, -67)
    pendown()
    fd(400)
    penup()
    setpos(-200, 66)
    pendown()
    fd(400)
    penup()
    seth(90)
    setpos(-67, -200)
    pendown()
    fd(400)
    penup()
    setpos(66, -200)
    pendown()
    fd(400)

def drawShape(s, x, y): #draws shape in grid box x and y = coord, s = shape type
    hideturtle()
    speed(100)
    pensize(6)
    penup()
    if s == 'X': #draws 'X'
        pencolor("orange")
        setpos(-266 (133*x), -266 (133*y))
        pendown()
        seth(135)
        fd(50)
        rt(180)
        fd(100)
        rt(180)
        fd(50)
        rt(90)
        fd(50)
        rt(180)
        fd(100) 
    elif s == 'O': #draws 'O'
        pencolor("green")
        setpos(-266 (133*x), -266 (133*y)-40)
        seth(0)
        pendown()
        circle(40)



def ai(): #EXPERIMENTAL AI
    x=0
    y=0
    d='O'
    e='O'
    f='O'
    for i in range(3): #checks positions
        if i == 0:
            d=' '
        elif i == 1:
            d='O'
            e=' '
        elif i == 2:
            d='O'
            e='O'
            f=' '
        for c in range(3):
            if score[c][0] == d and score[c][1] == e and score[c][2] == f:
                x = c 1
                y = i 1
                print('v',c)
            elif score[c][0] == d and score[c][1] == e and score[c][2] == f:
                x = i=1
                y = c 1
                print('h',c)
        if score[0][0] == d and score[1][1] == e and score[2][2] == f:
            print('lr',i)
            x = i 1
            y = i 1
        elif score[0][2] == d and score[1][1] == e and score[2][0] == f:
            print('rl',i)
            x = i 1
            y = 4-i
    d='X'
    e='X'
    f='X'
    if x == 0 and y == 0: #checks oposition positions
        for i in range(3):
            if i == 0:
                d=' '
            elif i == 1:
                d='X'
                e=' '
            elif i == 2:
                d='X'
                e='X'
                f=' '
            for c in range(3): 
                if score[c][0] == d and score[c][1] == e and score[c][2] == f:
                    x = c 1
                    y = i 1
                    print('op v')
                elif score[c][0] == d and score[c][1] == e and score[c][2] == f:
                    x = i=1
                    y = c 1
                    print('op v')
            if score[0][0] == d and score[1][1] == e and score[2][2] == f:
                x = i 1
                y = i 1
                print('op bt')
            elif score[0][2] == d and score[1][1] == e and score[2][0] == f:
                x = i 1
                y = 4-i
                print('op tb')

    if x == 0 and y == 0: #if no playable positions uses random
        x = randint(1,3)
        y = randint(1,3)
    return x, y

def valid(u,x,y): #checks player move is valid
    global attempts
    if x > 3 or y > 3:
        print ('Coordinate must be between 1 & 3')
    elif x == '' or y == '':
        print("Enter something!")
    elif score[y-1][x-1] == ' ':
        score[y-1][x-1] = u
        drawShape(u, x, y)
        attempts  =1
        return True
    elif score[y-1][x-1] == u:
        print("You've already gone here! ")
        return False
    elif score[y-1][x-1] != u:
        print("The other player is here! ")
        return False


def userAgent(u): #makes AI or user prompts and sets array
    global attempts
    global a
    global b
    if u == 0:
        a, b = ai()
        score[b-1][a-1] = 'O'
        print("The computer is taking its turn...")
        print(a,b)
        sleep(1)
        drawShape('O', a, b)
        attempts  =1
    else:
        x = input("Player " u ": enter x coordinate (1-3) ")
        y = input("Player " u ": enter y coordinate (1-3) ")
        try:
            x = int(x)
            y = int(y)
        except ValueError:
            print("That's not a valid number!")
            userAgent(u)
        while True:
            if valid(u,x,y) == True:
                break
            x = input("Player " u ": enter x coordinate (1-3) ")
            y = input("Player " u ": enter y coordinate (1-3) ")
            try:
                x = int(x)
                y = int(y)
            except ValueError:
                print("That's not a valid number!")


def checkWin(n): #checks for a player win (3 in row) or stalemate
    for i in range(3):
        if score[i][0] == n and score[i][1] == n and score[i][2] == n:
            print("Player " n " won!")
            return True
        elif score[0][i] == n and score[1][i] == n and score[2][i] == n:
            print("Player " n " won!")
            return True
    if score[0][0] == n and score[1][1] == n and score[2][2] == n:
        print("Player " n " won!")
        return True
    elif score[0][2] == n and score[1][1] == n and score[2][0] == n:
        print("Player " n " won!")
        return True
    elif attempts == 9:
        print("Stalemate!")
        return True
    else:
        return False

def printGrid():
    print(score[2])
    print(score[1])
    print(score[0])

from turtle import *
grid()
p = input("Are you playing by yourself? (SINGLE PLAYER EXPERIMENTAL) (y/n) ")

while True: #runs game until player win
    if p == 'y':
        userAgent('X')
        printGrid()
        if checkWin('X') == True:
            break
        userAgent(0)
        printGrid()
        if checkWin('O') == True:
            break

    elif p == 'n':
        userAgent('X')

        if checkWin('X') == True:
            break
        userAgent('O')

        if checkWin('O') == True:
            break

    else:
        print("You need to type y or n - try again!")
        p = input("Are you playing by yourself? (SINGLE PLAYER EXPERIMENTAL) (y/n) ")

input('Press ENTER to exit')

只是忽略AI功能,它现在已经永久性地实验(不起作用),但这不是问题。

我正在学校开放的晚上帮忙,我觉得如果两个人可以在不同的电脑上互相对战,那就太酷了。因此,我将在家用PC上托管服务器并进行端口转发,执行所有逻辑,客户端只需接收输入,将它们发送到服务器并由两个玩家绘制移动。我知道HTTP POST / GET,但是如何让服务器告诉客户端其他玩家移动了?我看过Twisted,看起来不错,但我真的不懂课程(我只是编程了一段时间)。

这将在学校的计算机上,因此我无法访问客户端的端口转发。理想情况下,我还有一个IP白名单,所以只有我想要的计算机才能访问服务器....

Soooo有谁可以帮助我在这里?我只需要了解服务器需要哪些代码以及客户端如何与之交互。谢谢:))

最佳答案:

3 个答案:

答案 0 :(得分:0)

让两个客户端通过服务器进行交互的最简单方法是让两个客户端轮询服务器以获取更新。 (你可以做服务器端事件,但这对于你正在寻找的东西来说太复杂了。)

但是,如果它们位于同一网络中,我建议在每个直接通信的设备上打开套接字服务器和客户端。

我会这样做:

  1. 在两台计算机上打开套接字服务器
  2. 启动游戏的人打开套接字客户端并将移动直接发送到服务器,然后重新打开套接字服务器
  3. 下次移动计算机会打开一个客户端并将其移回服务器。
  4. 这会删除所有运行服务器的不必要代码。

    有关套接字的信息,请参阅python手册。服务器/客户端底部都有很好的例子。

    https://docs.python.org/2/library/socket.html

    答案 1 :(得分:0)

    您可能想看看ZeroMQ。 http://zeromq.org/

    具体来说,你想要PyZMQ。 http://zeromq.github.io/pyzmq/api/zmq.html

    设置简单,您可以直接连接两台计算机。您可以在两者之间来回发送消息。

    绝对阅读文档,但您可能需要一个请求/回复模式。

    答案 2 :(得分:0)

    您可以从the twisted chatserver example

    开始
    """The most basic chat protocol possible.
    
    run me with twistd -y chatserver.py, and then connect with multiple
    telnet clients to port 1025
    """
    
    from twisted.protocols import basic
    
    
    
    class MyChat(basic.LineReceiver):
        def connectionMade(self):
            print "Got new client!"
            self.factory.clients.append(self)
    
        def connectionLost(self, reason):
            print "Lost a client!"
            self.factory.clients.remove(self)
    
        def lineReceived(self, line):
            print "received", repr(line)
            for c in self.factory.clients:
                c.message(line)
    
        def message(self, message):
            self.transport.write(message   '\n')
    
    
    from twisted.internet import protocol
    from twisted.application import service, internet
    
    factory = protocol.ServerFactory()
    factory.protocol = MyChat
    factory.clients = []
    
    application = service.Application("chatserver")
    internet.TCPServer(1025, factory).setServiceParent(application)
    

    客户端可以使用Python的内置telnetlib来连接

    如果您愿意,稍后您可以升级为使用twisted进行客户端连接。

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