如何在 Node.js 和 Express 4 上分离路由？

Question

我想从我的 server.js 文件中分离出路由。

我在 Scotch.io 上关注本教程 http://scotch.io/tutorials/javascript/build-a-restful-api-using-node-and-express-4

如果所有行都在 server.js 文件上，它就可以工作。但我无法分开。我怎样才能做到这一点？

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;

还有 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;

Answer 1

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;

*在 routs.js 中你应该定义中间件

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

Answer 2

就从主文件中分离路线而言..

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;

对于新项目，您也可以在命令行上尝试

express project_name

你需要快速生成器

Answer 3

另一种使用 Express 4.0 将路由分成自己的文件的方法：

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;
})();

Answer 4

一种将路由分离到自己的文件中的方法。

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
}

Answer 5

如果您将 express-4.x 与 TypeScript 和 ES6 一起使用，这将是最好的模板；

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");
});

使用var 和module.exports 更清晰可靠。

Answer 6

我遇到的一个问题是在使用 router.use 时尝试使用方法记录路径。使用最终使用此方法来解决它。允许您将路径保留到较高级别的较低路由器级别。

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;

如果您记录路由器堆栈，您实际上可以看到路径和方法

Answer 7

我们应该只需要 2 行代码

TL;DR

$ npm install express-routemagic --save

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

就是这样！

更多信息：

你会怎么做呢？让我们从文件结构开始：

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

请注意，在路线下有一个结构。 Route Magic 是文件夹感知的，并且会自动暗示这是您的 api uri 结构。

在 server.js 中

只需两行代码：

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

在路由/api/index.js中

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

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

在 routes/api/videos/index.js 中

Route Magic 知道您的文件夹结构并为您的 api 设置相同的结构，因此此 url 将是 api/videos

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

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

免责声明：我写了这个包。不过真的是姗姗来迟，等人来写已经到了我的极限了。

Answer 8

就我而言，我希望尽可能多地使用 Typescript。以下是我使用课程组织路线的方式：

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);
        })
    }
}

从您的 server.js 或您拥有服务器代码的地方，您可以通过以下方式调用 AuthService：

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) => {
    });
}