无法处理这个断开事件,不知道为什么套接字没有发送到客户端/客户端没有响应!
Server
io.sockets.on('connection', function (socket) {
socket.on('NewPlayer', function(data1) {
online = online + 1;
console.log('Online players : ' + online);
console.log('New player connected : ' + data1);
Players[data1] = data1;
console.log(Players);
});
socket.on('DelPlayer', function(data) {
delete Players[data];
console.log(Players);
console.log('Adios' + data);
});
socket.on('disconnect', function () {
socket.emit('disconnected');
online = online - 1;
});
});
Client
var socket = io.connect('http://localhost');
socket.on('connect', function () {
person_name = prompt("Welcome. Please enter your name");
socket.emit('NewPlayer', person_name);
socket.on('disconnected', function() {
socket.emit('DelPlayer', person_name);
});
});
正如您所看到的,当客户端断开连接时,数组对象[person_name] 应该被删除,但事实并非如此。
好的,不要通过名称跟踪玩家所连接的套接字来识别玩家。你可以有一个像这样的实现
Server
var allClients = [];
io.sockets.on('connection', function(socket) {
allClients.push(socket);
socket.on('disconnect', function() {
console.log('Got disconnect!');
var i = allClients.indexOf(socket);
allClients.splice(i, 1);
});
});
希望这能帮助您以另一种方式思考
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