How to separate routes on Node.js and Express 4?
node.jsExpressRouternode.js Problem Overview
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;
node.js Solutions
Solution 1 - node.js
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/
Solution 2 - node.js
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
Solution 3 - node.js
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;
})();
Solution 4 - node.js
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
}
Solution 5 - node.js
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.
Solution 6 - node.js
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
Solution 7 - node.js
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.
Solution 8 - node.js
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) => {
});
}