python - Pygame 和套接字在线

Question

您好，我从 tim online 的技术中了解了 pygame 在线多人游戏，我在他的第 5 课。这是这个的代码：Server.py

from _thread import *
from player import Player
import pickle

server = "192.168.1.5"
port = 5555

s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)

try:
    s.bind((server, port))
except socket.error as e:
    str(e)

s.listen(2)
print("Waiting for a connection, Server Started")


players = [Player(0,0,50,50,(255,0,0)), Player(100,100, 50,50, (0,0,255))]

def threaded_client(conn, player):
    conn.send(pickle.dumps(players[player]))
    reply = ""
    while True:
        try:
            data = pickle.loads(conn.recv(2048))
            players[player] = data

            if not data:
                print("Disconnected")
                break
            else:
                if player == 1:
                    reply = players[0]
                else:
                    reply = players[1]

                print("Received: ", data)
                print("Sending : ", reply)

            conn.sendall(pickle.dumps(reply))
        except:
            break

    print("Lost connection")
    conn.close()

currentPlayer = 0
while True:
    conn, addr = s.accept()
    print("Connected to:", addr)

    start_new_thread(threaded_client, (conn, currentPlayer))
    currentPlayer += 1

客户端.py：

import pygame
from network import Network
from player import Player

width = 500
height = 500
win = pygame.display.set_mode((width, height))
pygame.display.set_caption("Client")


def redrawWindow(win,player, player2):
    win.fill((255,255,255))
    player.draw(win)
    player2.draw(win)
    pygame.display.update()


def main():
    run = True
    n = Network()
    p = n.getP()
    clock = pygame.time.Clock()

    while run:
        clock.tick(60)
        p2 = n.send(p)

        for event in pygame.event.get():
            if event.type == pygame.QUIT:
                run = False
                pygame.quit()

        p.move()
        redrawWindow(win, p, p2)

main()

网络.py：

import socket
import pickle


class Network:
    def __init__(self):
        self.client = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
        self.server = "192.168.1.5"
        self.port = 5555
        self.addr = (self.server, self.port)
        self.p = self.connect()

    def getP(self):
        return self.p

    def connect(self):
        try:
            self.client.connect(self.addr)
            return pickle.loads(self.client.recv(2048))
        except:
            pass

    def send(self, data):
        try:
            self.client.send(pickle.dumps(data))
            return pickle.loads(self.client.recv(2048))
        except socket.error as e:
            print(e)

和 player.py：

import pygame

class Player():
    def __init__(self, x, y, width, height, color):
        self.x = x
        self.y = y
        self.width = width
        self.height = height
        self.color = color
        self.rect = (x,y,width,height)
        self.vel = 3

    def draw(self, win):
        pygame.draw.rect(win, self.color, self.rect)

    def move(self):
        keys = pygame.key.get_pressed()

        if keys[pygame.K_LEFT]:
            self.x -= self.vel

        if keys[pygame.K_RIGHT]:
            self.x += self.vel

        if keys[pygame.K_UP]:
            self.y -= self.vel

        if keys[pygame.K_DOWN]:
            self.y += self.vel

        self.update()

    def update(self):
        self.rect = (self.x, self.y, self.width, self.height)

好的，这就是代码。当我运行它时，我没有问题，但我只能运行 2 个客户端，如果有人关闭其中一个客户端，那么我将无法再次运行它。有什么解决办法吗？对不起，如果我的英语不好，因为我不太了解它。谢谢！

标签： pythonsocketspygamemultiplayer