javascript-如何在不复制代码的情况下删除耦合？

我正在为多人游戏制作服务器,但在弄清楚如何使逻辑模块化方面遇到一些问题.当玩家移动时,我想更新对象,更新数据库记录,并将移动发送给所有连接的客户端.现在,我正在这样做：

var socket = require('./socket');
var db = require('./db');

function Player(id, x, y) {
    this.id = id;
    this.x = x;
    this.y = y;
}

Player.prototype.move = function(x, y) {
    this.x = x;
    this.y = y;

    db.update({id: this.id}, {x: x, y: y});
    socket.emit('player moved', {x: x, y: y});
}

这将套接字和数据库紧密耦合到感觉错误的播放器对象.但是,我不想每次更新播放器对象时都在游戏循环中执行db.update和socket.emit.

正确的做法是什么？

解决方法:

我会做两件事：

1)让您的Player类发出事件.这样一来,您将来便可以添加更多触发器,而无需修改Player类

2)使用pos对象保存x和y值.这使代码更加简洁.

// In Player file
var EventEmitter = require('events').EventEmitter;

function Player(id, pos) {
  this.id = id;
  this.pos = pos;
}

Player.events = new EventEmitter();

Player.prototype.move = function (pos) {
  Player.events.emit('move', this);
};

module.exports = Player;

// In database file...
var db = require('./db');
var Player = require('./Player');

Player.events.on('move', function (player) {
  db.update({id: player.id}, player.pos);
});

// In socket file...
var socket = require('./socket');
var Player = require('./Player');

Player.events.on('move', function (player) {
  socket.emit('player moved', player.pos);
});

标签：node-js,game-engine,javascript
来源： https://codeday.me/bug/20191120/2042977.html