How to separate routes on Node.js and Express 4?

I want to separate Routes from my server.js file.

I am following this tutorial on Scotch.io http://scotch.io/tutorials/javascript/build-a-restful-api-using-node-and-express-4

It is working if all lines are on server.js file. But I am failing to separate. How can I make this work?

server.js

// set up ======================================================================
var express = require('express');
var app = express();
var bodyParser = require('body-parser');


// configuration ===============================================================
app.use(bodyParser());


var port = process.env.PORT || 8000;


var mongoose = require('mongoose');
var database = require('./config/database');
mongoose.connect(database.url);
var Video = require('./app/models/video');


// routes =======================================================================
app.use('/api', require('./app/routes/routes').router);


// listen (start app with node server.js) ======================================
app.listen(port);
console.log("ready captain, on deck" + port);


module.exports = app;

And the app/routes/routes.js

var express = require('express');
var router = express.Router();


router.use(function(req, res, next) {
console.log('Something is happening.');
next();
});


router.get('/', function(req, res) {
res.json({ message: 'hooray! welcome to our rest video api!' });
});




router.route('/videos')


.post(function(req, res) {


var video = new Video();
video.title = req.body.title;


video.save(function(err) {
if (err)
res.send(err);


res.json({ message: 'Video criado!' });
});




})


.get(function(req, res) {
Video.find(function(err, videos) {
if (err)
res.send(err);


res.json(videos);
});
});


module.exports.router = router;
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最佳答案

As far as separating routes from main file is concerned..

Server.js

//include the routes file
var routes = require('./routes/route');
var users = require('./routes/users');
var someapi = require('./routes/1/someapi');


////////
app.use('/', routes);
app.use('/users', users);
app.use('/1/someapi', someapi);

routes/route.js

//last line - try this
module.exports = router;

Also for new project you can try on command line

express project_name

You will need express-generator for that

小开

One way to separate routes into their own file.

SERVER.JS

var routes = require('./app/routes/routes');  //module you want to include
var app=express();
routes(app);   //routes shall use Express

ROUTES.JS

module.exports=function(app) {
//place your routes in here..
app.post('/api/..., function(req, res) {.....}   //example
}
小开

Another way to separate routes into their own files with Express 4.0:

server.js

var routes = require('./routes/routes');
app.use('/', routes);

routes.js

module.exports = (function() {
'use strict';
var router = require('express').Router();


router.get('/', function(req, res) {
res.json({'foo':'bar'});
});


return router;
})();
小开

Server.js

var express = require('express');
var app = express();


app.use(express.static('public'));


//Routes
app.use(require('./routes'));  //http://127.0.0.1:8000/    http://127.0.0.1:8000/about


//app.use("/user",require('./routes'));  //http://127.0.0.1:8000/user  http://127.0.0.1:8000/user/about




var server = app.listen(8000, function () {


var host = server.address().address
var port = server.address().port


console.log("Example app listening at http://%s:%s", host, port)


})

routes.js

var express = require('express');
var router = express.Router();


//Middle ware that is specific to this router
router.use(function timeLog(req, res, next) {
console.log('Time: ', Date.now());
next();
});




// Define the home page route
router.get('/', function(req, res) {
res.send('home page');
});


// Define the about route
router.get('/about', function(req, res) {
res.send('About us');
});




module.exports = router;

*In routs.js you should define Middle ware

ref http://wiki.workassis.com/nodejs-express-separate-routes/

小开

In my case, I like to have as much Typescript as possible. Here is how I organized my routes with classes:

export default class AuthService {
constructor() {
}


public login(): RequestHandler {
return this.loginUserFunc;
}


private loginUserFunc(req: Request, res: Response): void {
User.findOne({ email: req.body.email }, (err: any, user: IUser) => {
if (err)
throw err;
if(!user)
return res.status(403).send(AuthService.noSuccessObject());
else
return AuthService.comparePassword(user, req, res);
})
}
}

From your server.js or where you have your server code, you can call the AuthService in the following way:

import * as express from "express";
import AuthService from "./backend/services/AuthService";


export default class ServerApp {
private authService: AuthService;


this.authService = new AuthService();


this.myExpressServer.post("/api/login", this.authService.login(), (req: express.Request, res: express.Response) => {
});
}
小开

An issue I was running into was attempting to log the path with the methods when using router.use ended up using this method to resolve it. Allows you to keep path to a lower router level at the higher level.

routes.js

var express = require('express');
var router = express.Router();


var posts = require('./posts');


router.use(posts('/posts'));


module.exports = router;

posts.js

var express = require('express');
var router = express.Router();


let routeBuilder = path => {


router.get(`${path}`, (req, res) => {
res.send(`${path} is the path to posts`);
});


return router


}


module.exports = routeBuilder;

If you log the router stack you can actually see the paths and methods

小开

If you're using express-4.x with TypeScript and ES6, this would be a best template to use;

src/api/login.ts

import express, { Router, Request, Response } from "express";


const router: Router = express.Router();
// POST /user/signin
router.post('/signin', async (req: Request, res: Response) => {
try {
res.send('OK');
} catch (e) {
res.status(500).send(e.toString());
}
});


export default router;

src/app.ts

import express, { Request, Response } from "express";
import compression from "compression";  // compresses requests
import expressValidator from "express-validator";
import bodyParser from "body-parser";
import login from './api/login';


const app = express();


app.use(compression());
app.use(bodyParser.json());
app.use(bodyParser.urlencoded({ extended: true }));
app.use(expressValidator());


app.get('/public/hc', (req: Request, res: Response) => {
res.send('OK');
});


app.use('/user', login);


app.listen(8080, () => {
console.log("Press CTRL-C to stop\n");
});

Much clear and reliable rather using var and module.exports.

小开

We Ought To Only Need 2 Lines of Code

TL;DR

$ npm install express-routemagic --save

const magic = require('express-routemagic')
magic.use(app, __dirname, '[your route directory]')

That's it!

More info:

How you would do this? Let's start with file structuring:

project_folder
|--- routes
|     |--- api
|           |--- videos
|           |     |--- index.js
|           |
|           |--- index.js
|
|--- server.js

Note that under routes there is a structure. Route Magic is folder aware, and will imply this to be the api uri structure for you automatically.

In server.js

Just 2 lines of code:

const magic = require('express-routemagic')
magic.use(app, __dirname, 'routes')

In routes/api/index.js

const router = require('express').Router()


router.get('/', (req, res) => {
res.json({ message: 'hooray! welcome to our rest video api!' })
})

In routes/api/videos/index.js

Route Magic is aware of your folder structure and sets up the same structuring for your api, so this url will be api/videos

const router = require('express').Router()


router.post('/', (req, res) => { /* post the video */ })
router.get('/', (req, res) => { /* get the video */ })

Disclaimer: I wrote the package. But really it's long-overdue, it reached my limit to wait for someone to write it.

小开

I simply delcared the files and used require in the server.js file

 app.use(express.json());
require('./app/routes/devotion.route')(app);
require('./app/routes/user.route')(app);

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