Node.js와 Express 4에서 경로를 분리하는 방법은 무엇입니까?

Node.js와 Express 4에서 경로를 분리하는 방법은 무엇입니까?

---

내 server.js 파일에서 경로를 분리하고 싶습니다.

Scotch.io http://scotch.io/tutorials/javascript/build-a-restful-api-using-node-and-express-4 에서이 튜토리얼을 따르고 있습니다 .

모든 줄이 server.js 파일에 있으면 작동합니다. 그러나 나는 분리하지 못하고있다. 이 작업을 어떻게 할 수 있습니까?

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;

그리고 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;

---

주 파일에서 경로를 분리하는 한 ..

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);

route / route.js

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

또한 새 프로젝트의 경우 명령 줄에서 시도 할 수 있습니다.

express project_name

이를 위해서는 급행 발전기가 필요합니다

---

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)

})

route.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;

* routs.js에서 미들웨어를 정의해야합니다.

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

---

Express 4.0을 사용하여 경로를 자체 파일로 분리하는 또 다른 방법 :

server.js

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

route.js

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

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

    return router;
})();

---

경로를 자체 파일로 분리하는 한 가지 방법입니다.

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
}

---

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

---

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.

---

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) => {
    });
}

ReferenceURL : https://stackoverflow.com/questions/23923365/how-to-separate-routes-on-node-js-and-express-4