node.js require all files in a folder?

Question

How do I require all files in a folder in node.js?

need something like:

files.forEach(function (v,k){
  // require routes
  require('./routes/'+v);
}};

Answer 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.

yourfile.js

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

index.js

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

Answer 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 ) );
});

Answer 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.

Answer 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');

Answer 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 (!__hasProp.call(include, func)) continue;
      _results.push(exports[func] = include[func]);
    }
    return _results;
  }
};

fs.readdirSync('./fields/').forEach(collectExports);

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

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

Answer 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[reqModule.name] );
    }
  }
);

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

Answer 7

I know this question is 5+ years old, and the given answers are good, but I wanted something a bit more powerful for express, so i created the express-map2 package for npm. I was going to name it simply express-map, however the people at yahoo already have a package with that name, so i had to rename my package.

1. basic usage:

app.js (or whatever you call it)

var app = require('express'); // 1. include express

app.set('controllers',__dirname+'/controllers/');// 2. set path to your controllers.

require('express-map2')(app); // 3. patch map() into express

app.map({
    'GET /':'test',
    'GET /foo':'middleware.foo,test',
    'GET /bar':'middleware.bar,test'// seperate your handlers with a comma. 
});

controller usage:

//single function
module.exports = function(req,res){

};

//export an object with multiple functions.
module.exports = {

    foo: function(req,res){

    },

    bar: function(req,res){

    }

};

2. advanced usage, with prefixes:

app.map('/api/v1/books',{
    'GET /': 'books.list', // GET /api/v1/books
    'GET /:id': 'books.loadOne', // GET /api/v1/books/5
    'DELETE /:id': 'books.delete', // DELETE /api/v1/books/5
    'PUT /:id': 'books.update', // PUT /api/v1/books/5
    'POST /': 'books.create' // POST /api/v1/books
});

As you can see, this saves a ton of time and makes the routing of your application dead simple to write, maintain, and understand. it supports all of the http verbs that express supports, as well as the special .all() method.

• npm package: https://www.npmjs.com/package/express-map2
• github repo: https://github.com/r3wt/express-map

Answer 8

Expanding on this glob solution. Do this if you want to import all modules from a directory into index.js and then import that index.js in another part of the application. Note that template literals aren't supported by the highlighting engine used by stackoverflow so the code might look strange here.

const glob = require("glob");

let allOfThem = {};
glob.sync(`${__dirname}/*.js`).forEach((file) => {
  /* see note about this in example below */
  allOfThem = { ...allOfThem, ...require(file) };
});
module.exports = allOfThem;

Full Example

Directory structure

globExample/example.js
globExample/foobars/index.js
globExample/foobars/unexpected.js
globExample/foobars/barit.js
globExample/foobars/fooit.js

globExample/example.js

const { foo, bar, keepit } = require('./foobars/index');
const longStyle = require('./foobars/index');

console.log(foo()); // foo ran
console.log(bar()); // bar ran
console.log(keepit()); // keepit ran unexpected

console.log(longStyle.foo()); // foo ran
console.log(longStyle.bar()); // bar ran
console.log(longStyle.keepit()); // keepit ran unexpected

globExample/foobars/index.js

const glob = require("glob");
/*
Note the following style also works with multiple exports per file (barit.js example)
but will overwrite if you have 2 exports with the same
name (unexpected.js and barit.js have a keepit function) in the files being imported. As a result, this method is best used when
your exporting one module per file and use the filename to easily identify what is in it.

Also Note: This ignores itself (index.js) by default to prevent infinite loop.
*/

let allOfThem = {};
glob.sync(`${__dirname}/*.js`).forEach((file) => {
  allOfThem = { ...allOfThem, ...require(file) };
});

module.exports = allOfThem;

globExample/foobars/unexpected.js

exports.keepit = () => 'keepit ran unexpected';

globExample/foobars/barit.js

exports.bar = () => 'bar run';

exports.keepit = () => 'keepit ran';

globExample/foobars/fooit.js

exports.foo = () => 'foo ran';

From inside project with glob installed, run node example.js

$ node example.js
foo ran
bar run
keepit ran unexpected
foo ran
bar run
keepit ran unexpected

Answer 9

One module that I have been using for this exact use case is require-all.

It recursively requires all files in a given directory and its sub directories as long they don't match the excludeDirs property.

It also allows specifying a file filter and how to derive the keys of the returned hash from the filenames.

Answer 10

Require all files from routes folder and apply as middleware. No external modules needed.

// require
const { readdirSync } = require("fs");

// apply as middleware
readdirSync("./routes").map((r) => app.use("/api", require("./routes/" + r)));

Answer 11

I'm using node modules copy-to module to create a single file to require all the files in our NodeJS-based system.

The code for our utility file looks like this:

/**
 * Module dependencies.
 */

var copy = require('copy-to');
copy(require('./module1'))
.and(require('./module2'))
.and(require('./module3'))
.to(module.exports);

In all of the files, most functions are written as exports, like so:

exports.function1 = function () { // function contents };
exports.function2 = function () { // function contents };
exports.function3 = function () { // function contents };

So, then to use any function from a file, you just call:

var utility = require('./utility');

var response = utility.function2(); // or whatever the name of the function is

Answer 12

Using this function you can require a whole dir.

const GetAllModules = ( dirname ) => {
    if ( dirname ) {
        let dirItems = require( "fs" ).readdirSync( dirname );
        return dirItems.reduce( ( acc, value, index ) => {
            if ( PATH.extname( value ) == ".js" && value.toLowerCase() != "index.js" ) {
                let moduleName = value.replace( /.js/g, '' );
                acc[ moduleName ] = require( `${dirname}/${moduleName}` );
            }
            return acc;
        }, {} );
    }
}

// calling this function.

let dirModules = GetAllModules(__dirname);

Answer 13

Can use : https://www.npmjs.com/package/require-file-directory

• Require selected files with name only or all files.
• No need of absoulute path.
• Easy to understand and use.

Answer 14

Create an index.js file in your folder with this code :

const fs = require('fs')    
const files = fs.readdirSync('./routes')
for (const file of files) {
  require('./'+file)
}

And after that you can simply load all the folder with require("./routes")

Answer 15

If you include all files of *.js in directory example ("app/lib/*.js"):

In directory app/lib

example.js:

module.exports = function (example) { }

example-2.js:

module.exports = function (example2) { }

In directory app create index.js

index.js:

module.exports = require('./app/lib');