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requireDir()

Node helper to require() directories. The directory's files are examined, and each one that can be require()'d is require()'d and returned as part of a hash from that file's basename to its exported contents.

Example

Given this directory structure:

dir
+ a.js
+ b.json
+ c.coffee
+ d.txt

requireDir('./dir') will return the equivalent of:

{ a: require('./dir/a.js')
, b: require('./dir/b.json')
}

And if CoffeeScript has been registered via require('coffee-script/register'), c.coffee will also be returned.

Installation

npm install require-dir

Note that this package is not requireDir — turns out that's already taken! ;)

Usage

Basic usage that examines only directories' immediate files:

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

You can optionally customize the behavior by passing an extra options object:

var dir = requireDir('./path/to/dir', {recurse: true});

Options

recurse: Whether to recursively require() subdirectories too. (node_modules within subdirectories will be ignored.) Default is false.

camelcase: Automatically add camelcase aliases for files with dash- and underscore-separated names. E.g. foo-bar.js will be exposed under both the original 'foo-bar' name as well as a 'fooBar' alias. Default is false.

duplicates: By default, if multiple files share the same basename, only the highest priority one is require()'d and returned. (Priority is determined by the order of require.extensions keys, with directories taking precedence over files if recurse is true.) Specifying this option require()'s all files and returns full filename keys in addition to basename keys. Default is false.

E.g. in the example above, if there were also an a.json, the behavior would be the same by default, but specifying duplicates: true would yield:

{ a: require('./dir/a.js')
, 'a.js': require('./dir/a.js')
, 'a.json': require('./dir/a.json')
, b: require('./dir/b.json')
, 'b.json': require('./dir/b.json')
}

There might be more options in the future. ;)

Tips

If you want to require() the same directory in multiple places, you can do this in the directory itself! Just make an index.js file with the following:

module.exports = require('require-dir')();   // defaults to '.'

And don't worry, the calling file is always ignored to prevent infinite loops.

TODO

It'd be awesome if this could work with the regular require(), e.g. like a regular require() hook. Not sure that's possible though; directories are already special-cased to look for an index file or package.json.

An ignore option would be nice: a string or regex, or an array of either or both, of paths, relative to the directory, to ignore. String paths can be extensionless to ignore all extensions for that path. Supporting shell-style globs in string paths would be nice.

Currently, basenames are derived for directories too — e.g. a directory named a.txt will be returned as a when recursing — but should that be the case? Maybe directories should always be returned by their full name, and/or maybe this behavior should be customizable. This is hopefully an edge case.

License

MIT. © 2012-2015 Aseem Kishore.

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Node.js helper to require() directories.

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