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    3.0.0 • Public • Published


    An easy way to require all files within a directory.

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    var controllers = require('require-all')({
      dirname     :  __dirname + '/controllers',
      filter      :  /(.+Controller)\.js$/,
      excludeDirs :  /^\.(git|svn)$/,
      recursive   : true
    // controllers now is an object with references to all modules matching the filter
    // for example:
    // { HomeController: function HomeController() {...}, ...}

    Advanced usage

    If your objective is to simply require all .js and .json files in a directory you can just pass a string to require-all:

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

    Constructed object usage

    If your directory contains files that all export constructors, you can require them all and automatically construct the objects using resolve:

    var controllers = require('require-all')({
      dirname     :  __dirname + '/controllers',
      filter      :  /(.+Controller)\.js$/,
      resolve     : function (Controller) {
        return new Controller();

    Alternative property names

    If your directory contains files where the names do not match what you want in the resulting property (for example, you want camelCase but the file names are snake_case), then you can use the map function. The map function is called on both file and directory names, as they are added to the resulting object.

    var controllers = require('require-all')({
      dirname :  __dirname + '/controllers',
      filter  :  /(.+Controller)\.js$/,
      map     : function (name, path) {
        return name.replace(/_([a-z])/g, function (m, c) {
          return c.toUpperCase();

    Filtering files

    If your directory contains files that you do not want to require, or that you want only a part of the file's name to be used as the property name, filter can be a regular expression. In the following example, the filter is set to /^(.+Controller)\.js$/, which means only files that end in "Controller.js" are required, and the resulting property name will be the name of the file without the ".js" extension. For example, the file "MainController.js" will match, and since the first capture group will contain "MainController", that will be the property name used. If no capture group is used, then the entire match will be used as the name.

    var controllers = require('require-all')({
      dirname : __dirname + '/controllers',
      filter  : /^(.+Controller)\.js$/

    For even more advanced usage, the filter option also accepts a function that is invoked with the file name as the first argument. The filter function is expected to return a falsy value to ignore the file, otherwise a string to use as the property name.

    var controllers = requireAll({
      dirname : __dirname + '/controllers',
      filter  : function (fileName) {
        var parts = fileName.split('-');
        if (parts[1] !== 'Controller.js') return;
        return parts[0];

    Note that empty directories are always excluded from the end result.




    npm i require-all

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