您可以将所有路由函数放在其他文件(模块)中，并将其链接到主服务器文件。 在main express文件中，添加一个将模块链接到服务器的函数:

   function link_routes(app, route_collection){
       route_collection['get'].forEach(route => app.get(route.path, route.func));
       route_collection['post'].forEach(route => app.post(route.path, route.func));
       route_collection['delete'].forEach(route => app.delete(route.path, route.func));
       route_collection['put'].forEach(route => app.put(route.path, route.func));
   }

并为每个路由模型调用该函数:

link_routes(app, require('./login.js'))

在模块文件(例如- login.js文件)中，像往常一样定义函数:

const login_screen = (req, res) => {
    res.sendFile(`${__dirname}/pages/login.html`);
};

const forgot_password = (req, res) => {
    console.log('we will reset the password here')
}

然后用request方法导出它作为键，值是一个对象数组，每个对象都有路径和功能键。

module.exports = {
   get: [{path:'/',func:login_screen}, {...} ],
   post: [{path:'/login:forgotPassword', func:forgot_password}]
};   

如果你想把路由放在一个单独的文件中，例如routes.js，你可以这样创建routes.js文件:

module.exports = function(app){

    app.get('/login', function(req, res){
        res.render('login', {
            title: 'Express Login'
        });
    });

    //other routes..
}

然后你可以从app.js中以这样的方式传递app对象:

require('./routes')(app);

看看这些例子:https://github.com/visionmedia/express/tree/master/examples/route-separation

基于@ShadowCloud的例子，我能够在子目录中动态地包含所有路由。

路线/ index.js

var fs = require('fs');

module.exports = function(app){
    fs.readdirSync(__dirname).forEach(function(file) {
        if (file == "index.js") return;
        var name = file.substr(0, file.indexOf('.'));
        require('./' + name)(app);
    });
}

然后将路由文件放在routes目录中，如下所示:

路线/ test1.js

module.exports = function(app){

    app.get('/test1/', function(req, res){
        //...
    });

    //other routes..
}

根据需要重复多次，最后在app.js中放置

require('./routes')(app);

index.js

const express = require("express");
const app = express();
const http = require('http');
const server = http.createServer(app).listen(3000);
const router = (global.router = (express.Router()));
app.use('/books', require('./routes/books'))
app.use('/users', require('./routes/users'))
app.use(router);

路线/ users.js

const router = global.router
router.get('/', (req, res) => {
    res.jsonp({name: 'John Smith'})
}

module.exports = router

路线/ books.js

const router = global.router
router.get('/', (req, res) => {
    res.jsonp({name: 'Dreams from My Father by Barack Obama'})
}

module.exports = router

如果您的服务器在本地运行(http://localhost:3000)，那么

// Users
curl --request GET 'localhost:3000/users' => {name: 'John Smith'}

// Books
curl --request GET 'localhost:3000/books' => {name: 'Dreams from My Father by Barack Obama'}

我写了一个小插件来做这件事!我厌倦了一遍又一遍地写同样的代码。

https://www.npmjs.com/package/js-file-req

希望能有所帮助。

/routes文件夹中所有.js文件的完全递归路由，把它放在app.js中。

// Initialize ALL routes including subfolders
var fs = require('fs');
var path = require('path');

function recursiveRoutes(folderName) {
    fs.readdirSync(folderName).forEach(function(file) {

        var fullName = path.join(folderName, file);
        var stat = fs.lstatSync(fullName);

        if (stat.isDirectory()) {
            recursiveRoutes(fullName);
        } else if (file.toLowerCase().indexOf('.js')) {
            require('./' + fullName)(app);
            console.log("require('" + fullName + "')");
        }
    });
}
recursiveRoutes('routes'); // Initialize it

在/routes中你放入whatevername.js并像这样初始化你的路由:

module.exports = function(app) {
    app.get('/', function(req, res) {
        res.render('index', { title: 'index' });
    });

    app.get('/contactus', function(req, res) {
        res.render('contactus', { title: 'contactus' });
    });
}