CODEWITHSUNDEEP
node.jsexpressroutes
anonymousfox
anonymousfox

Reputation: 809

Dynamically load routes with express.js

I am using express.js as a webserver and would like an easy way to separate all the "app.get" and "app.post" functions to separate files. For example, if I would like to specify get and post functions for a login page, I would like to have a login.js file in a routes folder that is dynamically loaded (will automatically add all of the files without having to specify each one) when I run node app.js

I have tried this this solution!, but it isn't working for me.

Upvotes: 8

Views: 17048

Answers (4)

Aryan Vikash
Aryan Vikash

Reputation: 101

Typescript

routes/testroute.ts

import { Router } from 'express';

const router = Router();
router.get('/test',() => {
    // Do your stuffs Here
});



export = router;

index.ts

let app = express() 

const routePath = path.join(__dirname, 'routes');

fs.readdirSync(routePath).forEach(async (filename) => {
    let route = path.join(routePath, filename);
    try {
        const item = await import(route);
        app.use('/api', item.default);
    } catch (error) {
        console.log(error.message);
    }
});

app.listen()

Upvotes: 1

Julian
Julian

Reputation: 1467

With this approach, there is no need to write routes manually. Just setup a directory structure like the URL paths. Example route is at /routes/user/table/table.get.js and API route will be /user/table.

import app from './app'
import fs from 'fs-readdir-recursive'
import each from 'lodash/each'
import nth from 'lodash/nth'
import join from 'lodash/join'
import initial from 'lodash/initial'

const routes = fs(`${__dirname}/routes`)
each(routes, route => {
  let paths = route.split('/')

  // An entity has several HTTP verbs
  let entity = `/api/${join(initial(paths), '/')}`
  // The action contains a HTTP verb
  let action = nth(paths, -1)

  // Remove the last element to correctly apply action
  paths.pop()
  action = `./routes/${join(paths, '/')}/${action.slice(0, -3)}`

  app.use(entity, require(action))
})

Example route:

import { Router } from 'express'
import Table from '@models/table.model'

const routes = Router()

routes.get('/', (req, res, next) => {   
  Table
    .find({user: userIdentifier})
    .select('-user')
    .lean()
    .then(table => res.json(table))
    .catch(error => next(error))
})

module.exports = routes

Upvotes: -2

Paul Boomgaart
Paul Boomgaart

Reputation: 1

I ended up using a recursive approach to keep the code readable and asynchronous:

// routes     
processRoutePath(__dirname + "/routes");

function processRoutePath(route_path) {
    fs.readdirSync(route_path).forEach(function(file) {
        var filepath = route_path + '/' + file;
        fs.stat(filepath, function(err,stat) {
            if (stat.isDirectory()) {
                processRoutePath(filepath);
            } else {
                console.info('Loading route: ' + filepath);
                require(filepath)(app, passport);
            }
        });
    });
}

This could be made more robust by checking fro correct file extensions etc, but I keep my routes folder clean and did not want the added complexity

Upvotes: 0

sachin
sachin

Reputation: 14375

app.js

var express=require("express");
var app=express();
var fs=require("fs");
var routePath="./routers/"; //add one folder then put your route files there my router folder name is routers
fs.readdirSync(routePath).forEach(function(file) {
    var route=routePath+file;
    require(route)(app);
});
app.listen(9123);

I have put below two routers in that folder

route1.js

module.exports=function(app){
  app.get('/',function(req,res){
     res.send('/ called successfully...');
  });
}

route2.js

module.exports=function(app){
app.get('/upload',function(req,res){
  res.send('/upload called successfully...');
});
}

Upvotes: 19

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