node.js require all files in a folder?

node.js require all files in a folder?

How do I require all files in a folder in node.js?
need something like:
files.forEach(function (v,k){
// require routes


Solution 1:

When require is given the path of a folder, it’ll look for an index.js file in that folder; if there is one, it uses that, and if there isn’t, it fails.

It would probably make most sense (if you have control over the folder) to create an index.js file and then assign all the “modules” and then simply require that.


var routes = require("./routes");


exports.something = require("./routes/something.js");
exports.others = require("./routes/others.js");

If you don’t know the filenames you should write some kind of loader.

Working example of a loader:

var normalizedPath = require("path").join(__dirname, "routes");

require("fs").readdirSync(normalizedPath).forEach(function(file) {
  require("./routes/" + file);

// Continue application logic here

Solution 2:

I recommend using glob to accomplish that task.

var glob = require( 'glob' )
  , path = require( 'path' );

glob.sync( './routes/**/*.js' ).forEach( function( file ) {
  require( path.resolve( file ) );

Solution 3:

Base on @tbranyen’s solution, I create an index.js file that load arbitrary javascripts under current folder as part of the exports.

// Load `*.js` under current directory as properties
//  i.e., `User.js` will become `exports['User']` or `exports.User`
require('fs').readdirSync(__dirname + '/').forEach(function(file) {
  if (file.match(/\.js$/) !== null && file !== 'index.js') {
    var name = file.replace('.js', '');
    exports[name] = require('./' + file);

Then you can require this directory from any where else.

Solution 4:

Another option is to use the package require-dir which let’s you do the following. It supports recursion as well.

var requireDir = require('require-dir');
var dir = requireDir('./path/to/dir');

Solution 5:

I have a folder /fields full of files with a single class each, ex:

fields/Text.js -> Test class
fields/Checkbox.js -> Checkbox class

Drop this in fields/index.js to export each class:

var collectExports, fs, path,
  __hasProp = {}.hasOwnProperty;

fs = require('fs');    
path = require('path');

collectExports = function(file) {
  var func, include, _results;

  if (path.extname(file) === '.js' && file !== 'index.js') {
    include = require('./' + file);
    _results = [];
    for (func in include) {
      if (!, func)) continue;
      _results.push(exports[func] = include[func]);
    return _results;


This makes the modules act more like they would in Python:

var text = new Fields.Text()
var checkbox = new Fields.Checkbox()

Solution 6:

One more option is require-dir-all combining features from most popular packages.

Most popular require-dir does not have options to filter the files/dirs and does not have map function (see below), but uses small trick to find module’s current path.

Second by popularity require-all has regexp filtering and preprocessing, but lacks relative path, so you need to use __dirname (this has pros and contras) like:

var libs = require('require-all')(__dirname + '/lib');

Mentioned here require-index is quite minimalistic.

With map you may do some preprocessing, like create objects and pass config values (assuming modules below exports constructors):

// Store config for each module in config object properties 
// with property names corresponding to module names 
var config = {
  module1: { value: 'config1' },
  module2: { value: 'config2' }

// Require all files in modules subdirectory 
var modules = require('require-dir-all')(
  'modules', // Directory to require 
  { // Options 
    // function to be post-processed over exported object for each require'd module 
    map: function(reqModule) {
      // create new object with corresponding config passed to constructor 
      reqModule.exports = new reqModule.exports( config[] );

// Now `modules` object holds not exported constructors, 
// but objects constructed using values provided in `config`.