Question

我目前正在尝试使用express 4.x和socket io 1.4实现会话管理，引用this answer。问题是快速会话函数的第二个参数是res（响应）对象，它在我的路由中返回'undefined'。这个问题已经过时了还是我做错了什么？

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这是我的sockets.js

var http = require('http');

var session = require('express-session')(
{
  saveUninitialized : false,
  resave:false,
  secret:'secretstuff',
  cookie : {
    path : '/'
  }
}
);

var express = require('express');

var app = express();

app.use(session)

var http_server = http.createServer(app);

var io = require('./sockets')(http_server,session);

这是我得到的地方

未捕获的TypeError：参数res是必需的

Answer 1

在许多资料中提出的

sessionMiddleware()函数对我来说效果很好：

io.use(function(socket, next) {
    sessionMiddleware(socket.request, socket.request.res, next);
});

使用网络socket.io-client <=> socket.io服务器，直到我将socket.io-client-cpp应用程序添加到链中-服务器在socket.io-client-cpp连接时崩溃，并出现上述错误：

TypeError: argument res is required

在sessionMiddleware()函数中

。人们建议从中间件中完全删除socket.request.res：

，并替换为{}。但是，认为如果有人仍然需要并使用socket.request.res，可以将其更改为稍微更好的变体：

io.use(function(socket, next) {
    sessionMiddleware(socket.request, socket.request.res || {}, next);
});

这对我来说很好！

使用快速会话的套接字IO - socket.request.res undefined

1 个答案: