Home » Nodejs » node.js require from parent folder

node.js require from parent folder

Posted by: admin December 21, 2017 Leave a comment


I have the following structure:

-- node_modules
-- websites
---- common
------- config.js
---- testing
------- test.js

Inside config I have some variables set, which are being exported using module.export.

I am trying to retrieve those variables when running node test.js from config.js using the following codes:

var configData = require('./common/config.js')
var configData = require('../common/config.js')

None of them work. What can I do to retrieve the data from the other folder?

var configData = require('./../common/config.js');
  1. ./ is testing/

  2. ./../ is websites/

  3. ./../common/ is websites/common/

  4. ./../common/config.js is websites/common/config.js


from test.js:

const configData = require('../common/config');

You can safely omit '.js'.

As documentation say:

File Modules

If the exact filename is not found, then Node.js will attempt to load the required filename with the added extensions: .js,
.json, and finally .node.

.js files are interpreted as JavaScript text files, and .json files
are parsed as JSON text files. .node files are interpreted as compiled
addon modules loaded with dlopen.

A required module prefixed with ‘/’ is an absolute path to the file.
For example, require(‘/home/marco/foo.js’) will load the file at

A required module prefixed with ‘./’ is relative to the file calling
. That is, circle.js must be in the same directory as foo.js
for require(‘./circle’) to find it.

Without a leading ‘/’, ‘./’, or ‘../’ to indicate a file, the module
must either be a core module or is loaded from a node_modules folder

If the given path does not exist, require() will throw an Error with
its code property set to ‘MODULE_NOT_FOUND’.

More info about how require() work here.