How can I make Node.js 'require' absolute? (instead of relative)

Question

I would like to 'require' my files always by the root of my project and not relative to the current module.

For example, if you look at Express.js' app.js line 6, you will see

express = require('../../')

That's really bad, IMO. Imagine I would like to put all my examples closer to the root only by one level. That would be impossible, because I would have to update more than 30 examples and many times within each example. To this:

express = require('../')

My solution would be to have a special case for root based: if a string starts with an $ then it's relative to the root folder of the project.

What can I do?

Update 2

Now I'm using RequireJS which allows you to write in one way and works both on client and on server. RequireJS also allows you to create custom paths.

Update 3

Now I moved to Webpack and Gulp.js and I use enhanced-require to handle modules on the server side. See here for the rationale: http://hackhat.com/p/110/module-loader-webpack-vs-requirejs-vs-browserify/

Answer 1

Use:

var myModule = require.main.require('./path/to/module');

It requires the file as if it were required from the main JavaScript file, so it works pretty well as long as your main JavaScript file is at the root of your project... and that's something I appreciate.

Answer 2

There's a really interesting section in the Browserify Handbook:

avoiding ../../../../../../..

Not everything in an application properly belongs on the public npm and the overhead of setting up a private npm or git repo is still rather large in many cases. Here are some approaches for avoiding the ../../../../../../../ relative paths problem.

node_modules

People sometimes object to putting application-specific modules into node_modules because it is not obvious how to check in your internal modules without also checking in third-party modules from npm.

The answer is quite simple! If you have a .gitignore file that ignores node_modules:

node_modules

You can just add an exception with ! for each of your internal application modules:

node_modules/*
!node_modules/foo
!node_modules/bar

Please note that you can't unignore a subdirectory, if the parent is already ignored. So instead of ignoring node_modules, you have to ignore every directory inside node_modules with the node_modules/* trick, and then you can add your exceptions.

Now anywhere in your application you will be able to require('foo') or require('bar') without having a very large and fragile relative path.

If you have a lot of modules and want to keep them more separate from the third-party modules installed by npm, you can just put them all under a directory in node_modules such as node_modules/app:

node_modules/app/foo
node_modules/app/bar

Now you will be able to require('app/foo') or require('app/bar') from anywhere in your application.

In your .gitignore, just add an exception for node_modules/app:

node_modules/*
!node_modules/app

If your application had transforms configured in package.json, you'll need to create a separate package.json with its own transform field in your node_modules/foo or node_modules/app/foo component directory because transforms don't apply across module boundaries. This will make your modules more robust against configuration changes in your application and it will be easier to independently reuse the packages outside of your application.

symlink

Another handy trick if you are working on an application where you can make symlinks and don't need to support windows is to symlink a lib/ or app/ folder into node_modules. From the project root, do:

ln -s ../lib node_modules/app

and now from anywhere in your project you'll be able to require files in lib/ by doing require('app/foo.js') to get lib/foo.js.

custom paths

You might see some places talk about using the $NODE_PATH environment variable or opts.paths to add directories for node and browserify to look in to find modules.

Unlike most other platforms, using a shell-style array of path directories with $NODE_PATH is not as favorable in node compared to making effective use of the node_modules directory.

This is because your application is more tightly coupled to a runtime environment configuration so there are more moving parts and your application will only work when your environment is setup correctly.

node and browserify both support but discourage the use of $NODE_PATH.

Answer 3

I like to make a new node_modules folder for shared code. Then let Node.js and 'require' do what they do best.

For example:

- node_modules // => these are loaded from your *package.json* file
- app
  - node_modules // => add node-style modules
    - helper.js
  - models
    - user
    - car
- package.json
- .gitignore

For example, if you're in car/index.js you can require('helper') and Node.js will find it!

How node_modules Work

Node.js has a clever algorithm for resolving modules that is unique among rival platforms.

If you require('./foo.js') from /beep/boop/bar.js, Node.js will look for ./foo.js in /beep/boop/foo.js. Paths that start with a ./ or ../ are always local to the file that calls require().

If, however, you 'require' a non-relative name such as require('xyz') from /beep/boop/foo.js, Node.js searches these paths in order, stopping at the first match and raising an error if nothing is found:

/beep/boop/node_modules/xyz
/beep/node_modules/xyz
/node_modules/xyz

For each xyz directory that exists, Node.js will first look for a xyz/package.json to see if a "main" field exists. The "main" field defines which file should take charge if you require() the directory path.

For example, if /beep/node_modules/xyz is the first match and /beep/node_modules/xyz/package.json has:

{
  "name": "xyz",
  "version": "1.2.3",
  "main": "lib/abc.js"
}

then the exports from /beep/node_modules/xyz/lib/abc.js will be returned by require('xyz').

If there is no package.json or no "main" field, index.js is assumed:

/beep/node_modules/xyz/index.js

Answer 4

The big picture

It seems "really bad" but give it time. It is, in fact, really good. The explicit require()s give a total transparency and ease of understanding that is like a breath of fresh air during a project life cycle.

Think of it this way: You are reading an example, dipping your toes into Node.js and you've decided it is "really bad IMO." You are second-guessing leaders of the Node.js community, people who have logged more hours writing and maintaining Node.js applications than anyone. What is the chance the author made such a rookie mistake? (And I agree, from my Ruby and Python background, it seems at first like a disaster.)

There is a lot of hype and counter-hype surrounding Node.js. But when the dust settles, we will acknowledge that explicit modules and "local first" packages were a major driver of adoption.

The common case

Of course, node_modules from the current directory, then the parent, then grandparent, great-grandparent, etc. is searched. So packages you have installed already work this way. Usually you can require("express") from anywhere in your project and it works fine.

If you find yourself loading common files from the root of your project (perhaps because they are common utility functions), then that is a big clue that it's time to make a package. Packages are very simple: move your files into node_modules/ and put a package.json there. Voila! Everything in that namespace is accessible from your entire project. Packages are the correct way to get your code into a global namespace.

Other workarounds

I personally don't use these techniques, but they do answer your question, and of course you know your own situation better than I.

You can set $NODE_PATH to your project root. That directory will be searched when you require().

Next, you could compromise and require a common, local file from all your examples. That common file simply re-exports the true file in the grandparent directory.

examples/downloads/app.js (and many others like it)

var express = require('./express')

examples/downloads/express.js

module.exports = require('../../')

Now when you relocate those files, the worst-case is fixing the one shim module.

Answer 5

If you are using yarn instead of npm you can use workspaces.

Let's say I have a folder services I wish to require more easily:

.
├── app.js
├── node_modules
├── test
├── services
│   ├── foo
│   └── bar
└── package.json

To create a Yarn workspace, create a package.json file inside the services folder:

{
  "name": "myservices",
  "version": "1.0.0"
}

In your main package.json add:

"private": true,
"workspaces": ["myservices"]

Run yarn install from the root of the project.

Then, anywhere in your code, you can do:

const { myFunc } = require('myservices/foo')

instead of something like:

const { myFunc } = require('../../../../../../services/foo')

Answer 6

Have a look at node-rfr.

It's as simple as this:

var rfr = require('rfr');
var myModule = rfr('projectSubDir/myModule');

Answer 7

I use process.cwd() in my projects. For example:

var Foo = require(process.cwd() + '/common/foo.js');

It might be worth noting that this will result in requireing an absolute path, though I have yet to run into issues with this.

Answer 8

IMHO, the easiest way is to define your own function as part of GLOBAL object. Create projRequire.js in the root of you project with the following contents:

var projectDir = __dirname;

module.exports = GLOBAL.projRequire = function(module) {
  return require(projectDir + module);
}

In your main file before requireing any of project-specific modules:

// init projRequire
require('./projRequire');

After that following works for me:

// main file
projRequire('/lib/lol');

// index.js at projectDir/lib/lol/index.js
console.log('Ok');

---

@Totty, I've comed up with another solution, which could work for case you described in comments. Description gonna be tl;dr, so I better show a picture with structure of my test project.

Answer 9

Assuming your project root is the current working directory, this should work:

// require built-in path module
path = require('path');

// require file relative to current working directory
config = require( path.resolve('.','config.js') );

Answer 10

There's a good discussion of this issue here.

I ran into the same architectural problem: wanting a way of giving my application more organization and internal namespaces, without:

• mixing application modules with external dependencies or bothering with private npm repos for application-specific code
• using relative requires, which make refactoring and comprehension harder
• using symlinks or changing the node path, which can obscure source locations and don't play nicely with source control

In the end, I decided to organize my code using file naming conventions rather than directories. A structure would look something like:

• npm-shrinkwrap.json
• package.json
• node_modules
  • ...
• src
  • app.js
  • app.config.js
  • app.models.bar.js
  • app.models.foo.js
  • app.web.js
  • app.web.routes.js
  • ...

Then in code:

var app_config = require('./app.config');
var app_models_foo = require('./app.models.foo');

or just

var config = require('./app.config');
var foo = require('./app.models.foo');

and external dependencies are available from node_modules as usual:

var express = require('express');

In this way, all application code is hierarchically organized into modules and available to all other code relative to the application root.

The main disadvantage is of course that in a file browser, you can't expand/collapse the tree as though it was actually organized into directories. But I like that it's very explicit about where all code is coming from, and it doesn't use any 'magic'.

Answer 11

I have tried many of these solutions. I ended up adding this to the top of my main file (e.g. index.js):

process.env.NODE_PATH = __dirname;
require('module').Module._initPaths();

This adds the project root to the NODE_PATH when the script is loaded. The allows me to require any file in my project by referencing its relative path from the project root such as var User = require('models/user'). This solution should work as long as you are running a main script in the project root before running anything else in your project.

Answer 12

Some of the answers are saying that the best way is to add the code to the node_modules folder folder as a package. I agree and it's probably the best way to lose the ../../../ in require, but none of them actually give a way to do so.

From version 2.0.0, you can install a package from local files, which means you can create a folder in your root with all the packages you want,

-modules
 --foo
 --bar
-app.js
-package.json

so in package.json you can add the modules (or foo and bar) as a package without publishing or using an external server like this:

{
  "name": "baz",
  "dependencies": {
    "bar": "file: ./modules/bar",
    "foo": "file: ./modules/foo"
  }
}

After that you do npm install, and you can access the code with var foo = require("foo"), just like you do with all the other packages.

More information can be found on package.json.

And here is how to create a package: Creating Node.js modules

Answer 13

You could use a module I made, Undot. It is nothing advanced, just a helper so you can avoid those dot hell with simplicity.

Example:

var undot = require('undot');
var User = undot('models/user');
var config = undot('config');
var test = undot('test/api/user/auth');

Answer 14

Another answer:

Imagine this folders structure:

• node_modules - lodash

• src - subdir - foo.js - bar.js - main.js

• tests

  - test.js
  

Then in test.js, you need to require files like this:

const foo = require("../src/subdir/foo");
const bar = require("../src/subdir/bar");
const main = require("../src/main");
const _ = require("lodash");

and in main.js:

const foo = require("./subdir/foo");
const bar = require("./subdir/bar");
const _ = require("lodash");

Now you can use babel and the babel-plugin-module-resolver with this .babelrc file to configure two root folders:

{
    "plugins": [
        ["module-resolver", {
            "root": ["./src", "./src/subdir"]
        }]
    ]
}

Now you can require files in the same manner in tests and in src:

const foo = require("foo");
const bar = require("bar");
const main = require("main");
const _ = require("lodash");

And if you want use the ES6 module syntax:

{
    "plugins": [
        ["module-resolver", {
            "root": ["./src", "./src/subdir"]
        }],
        "transform-es2015-modules-commonjs"
    ]
}

then you import files in tests and src like this:

import foo from "foo"
import bar from "bar"
import _ from "lodash"

Answer 15

Manual Symlinks (and Windows Junctions)

Couldn't the examples directory contain a node_modules with a symbolic link to the root of the project project -> ../../ thus allowing the examples to use require('project'), although this doesn't remove the mapping, it does allow the source to use require('project') rather than require('../../').

I have tested this, and it does work with v0.6.18.

Listing of project directory:

$ ls -lR project
project:
drwxr-xr-x 3 user user 4096 2012-06-02 03:51 examples
-rw-r--r-- 1 user user   49 2012-06-02 03:51 index.js

project/examples:
drwxr-xr-x 2 user user 4096 2012-06-02 03:50 node_modules
-rw-r--r-- 1 user user   20 2012-06-02 03:51 test.js

project/examples/node_modules:
lrwxrwxrwx 1 user user 6 2012-06-02 03:50 project -> ../../

The contents of index.js assigns a value to a property of the exports object and invokes console.log with a message that states it was required. The contents of test.js is require('project').

Automated Symlinks

The problem with manually creating symlinks is that every time you npm ci, you lose the symlink. If you make the symlink process a dependency, viola, no problems.

The module basetag is a postinstall script that creates a symlink (or Windows junction) named $ every time npm install or npm ci is run:

npm install --save basetag

node_modules/$ -> ..

With that, you don't need any special modification to your code or require system. $ becomes the root from which you can require.

var foo = require('$/lib/foo.js');

If you don't like the use of $ and would prefer # or something else (except @, which is a special character for npm), you could fork it and make the change.

Note: Although Windows symlinks (to files) require admin permissions, Windows junctions (to directories) do not need Windows admin permissions. This is a safe, reliable, cross-platform solution.

Answer 16

You could define something like this in your app.js:

requireFromRoot = (function(root) {
    return function(resource) {
        return require(root+"/"+resource);
    }
})(__dirname);

and then anytime you want to require something from the root, no matter where you are, you just use requireFromRoot instead of the vanilla require. Works pretty well for me so far.

Answer 17

Imho the easiest way to achieve this is by creating a symbolic link on app startup at node_modules/app (or whatever you call it) which points to ../app. Then you can just call require("app/my/module"). Symbolic links are available on all major platforms.

However, you should still split your stuff in smaller, maintainable modules which are installed via npm. You can also install your private modules via git-url, so there is no reason to have one, monolithic app-directory.

Answer 18

In your own project you could modify any .js file that is used in the root directory and add its path to a property of the process.env variable. For example:

// in index.js
process.env.root = __dirname;

Afterwards you can access the property everywhere:

// in app.js
express = require(process.env.root);

Answer 19

Here is the actual way I'm doing for more than 6 months. I use a folder named node_modules as my root folder in the project, in this way it will always look for that folder from everywhere I call an absolute require:

• node_modules
  • myProject
    • index.js I can require("myProject/someFolder/hey.js") instead of require("./someFolder/hey.js")
    • someFolder which contains hey.js

This is more useful when you are nested into folders and it's a lot less work to change a file location if is set in absolute way. I only use 2 the relative require in my whole app.

Answer 20

I just came across this article which mentions app-module-path. It allows you to configure a base like this:

require('app-module-path').addPath(baseDir);

Answer 21

I was looking for the exact same simplicity to require files from any level and I found module-alias.

Just install:

npm i --save module-alias

Open your package.json file, here you can add aliases for your paths, for e.g.

"_moduleAliases": {
 "@root"      : ".", // Application's root
 "@deep"      : "src/some/very/deep/directory/or/file",
 "@my_module" : "lib/some-file.js",
 "something"  : "src/foo", // Or without @. Actually, it could be any string
}

And use your aliases by simply:

require('module-alias/register')
const deep = require('@deep')
const module = require('something')

Answer 22

If anyone's looking for yet another way to get around this problem, here's my own contribution to the effort:

https://www.npmjs.com/package/use-import

The basic idea: you create a JSON file in the root of the project that maps your filepaths to shorthand names (or get use-automapper to do it for you). You can then request your files/modules using those names. Like so:

var use = require('use-import');
var MyClass = use('MyClass');

So there's that.

Answer 23

What I like to do is leverage how Node.js loads from the node_modules directory for this.

If one tries to load the module "thing", one would do something like

require('thing');

Node.js will then look for the 'thing' directory in the 'node_modules' directory.

Since the node_modules folder is normally at the root of the project, we can leverage this consistency. (If node_modules is not at the root, then you have other self-induced headaches to deal with.)

If we go into the directory and then back out of it, we can get a consistent path to the root of the Node.js project.

require('thing/../../');

Then if we want to access the /happy directory, we would do this:

require('thing/../../happy');

Though it is quite a bit hacky, however I feel if the functionality of how node_modules load changes, there will be bigger problems to deal with. This behavior should remain consistent.

To make things clear, I do this, because the name of module does not matter.

require('root/../../happy');

I used it recently for Angular 2. I want to load a service from the root.

import {MyService} from 'root/../../app/services/http/my.service';

Answer 24

I wrote this small package that lets you require packages by their relative path from project root, without introducing any global variables or overriding node defaults

https://github.com/Gaafar/pkg-require

It works like this

// create an instance that will find the nearest parent dir containing package.json from your __dirname
const pkgRequire = require('pkg-require')(__dirname);

// require a file relative to the your package.json directory 
const foo = pkgRequire('foo/foo')

// get the absolute path for a file
const absolutePathToFoo = pkgRequire.resolve('foo/foo')

// get the absolute path to your root directory
const packageRootPath = pkgRequire.root()

Answer 25

I just want to follow up on the great answer from Paolo Moretti and Browserify. If you are using a transpiler (e.g., babel, typescript) and you have separate folders for source and transpiled code like src/ and dist/, you could use a variation of the solutions as

node_modules

With the following directory structure:

app
  node_modules
    ... // Normal npm dependencies for app
  src
    node_modules
      app
        ... // Source code
  dist
    node_modules
      app
        ... // Transpiled code

You can then let Babel, etc. to transpile the src directory to the dist directory.

Symbolic link

Using a symbolic link, we can get rid some levels of nesting:

app
  node_modules
    ... // Normal npm dependencies for app
  src
    node_modules
      app // Symbolic links to '..'
    ... // Source code
  dist
    node_modules
      app // Symbolic links to '..'
    ... // Transpiled code

A caveat with babel --copy-files: The --copy-files flag of babel does not deal with symbolic links well. It may keep navigating into the .. symlink and recursively seeing endless files. A workaround is to use the following directory structure:

app
  node_modules
    app // Symbolic link to '../src'
    ... // Normal npm dependencies for app
  src
    ... // Source code
  dist
    node_modules
      app // Symbolic links to '..'
    ... // Transpiled code

In this way, code under src will still have app resolved to src, whereas Babel would not see symlinks anymore.

Answer 26

I had the same problem many times. This can be solved by using the basetag npm package. It doesn't have to be required itself, only installed as it creates a symlink inside node_modules to your base path.

const localFile = require('$/local/file')
// instead of
const localFile = require('../../local/file')

Using the $/... prefix will always reference files relative to your apps root directory.

Source: How I created basetag to solve this problem

Answer 27

Using the "Imports" property

As of March 2023, a good way to eliminate the NodeJS relative paths is to use the imports property in package.json. For more information, please refer to this post:

• How to import JavaScript files in NodeJS using file paths from project root?.

In the codes below, #root is the project root.

(Please kindly upvote this answer and this post if they help you. Thanks!)

For CommonJS-style JavaScripts:

// package.json
{
  "imports": {
    "#root/*.js": "./*.js"
  }
}

// main.js:
const Source = require('#root/path/to/Source.js');

// Source.js:
module.exports = class Source {
  // ...
}

For ECMAScript-style JavaScripts:

// package.json:
{
  "type" : "module",
  "imports": {
    "#root/*.js": "./*.js"
  }
}

// main.js
import { Source } from '#root/path/to/Source.js';

// Source.js:
export class Source {
  // ...
}

Advantages:

• No need to "import" or "require" any additional packages (No Babel.js, No Webpack, No RequireJS). After installing NodeJS, this method works out of the box.

• IDE linkages work as expected (Ctrl-click a class name to jump directly to the source file. Also, moving the source file (by drag and drop) will automatically update the file path references. Tested on WebStorm 2022.3.2 and VS Code 1.76.2.)

• Works with both .mjs (ECMAScript module system) and .cjs (CommonJS) file types. Please see this reference Post on .cjs and .mjs.

• No need to modify with the reserved node_modules directory

• No need to set up any linux file links at the OS level

Answer 28

I created a node module called rekuire.

It allows you to 'require' without the use of relative paths.

It is super easy to use.

Answer 29

We are about to try a new way to tackle this problem.

Taking examples from other known projects like Spring Framework and Guice, we will define a "context" object which will contain all the "require" statement.

This object will then be passed to all other modules for use.

For example,

var context = {}

context.module1 = require("./module1")( { "context" : context } )
context.module2 = require("./module2")( { "context" : context } )

This requires us to write each module as a function that receives opts, which looks to us as a best practice anyway...

module.exports = function(context){ ... }

And then you will refer to the context instead of requiring stuff.

var module1Ref = context.moduel1;

If you want to, you can easily write a loop to do the 'require' statements

var context = {};
var beans = {"module1" : "./module1","module2" : "./module2" };
for ( var i in beans ){
    if ( beans.hasOwnProperty(i)){
         context[i] = require(beans[i])(context);
    }
};

This should make life easier when you want to mock (tests) and also solves your problem along the way while making your code reusable as a package.

You can also reuse the context initialization code by separating the beans declaration from it.

For example, your main.js file could look like so

var beans = { ... }; // like before
var context = require("context")(beans); // This example assumes context is a node_module since it is reused..

This method also applies to external libraries, and there isn't any need to hard code their names every time we require them. However, it will require a special treatment as their exports are not functions that expect context...

Later on, we can also define beans as functions—which will allow us to require different modules according to the environment—but that it out of this question's scope.

Answer 30

I was having trouble with this same issue, so I wrote a package called include.

Include handles figuring out your project's root folder by way of locating your package.json file, then passes the path argument you give it to the native require() without all of the relative path mess. I imagine this not as a replacement for require(), but a tool for requiring handling non-packaged / non-third-party files or libraries. Something like

var async = require('async'),
    foo   = include('lib/path/to/foo')

Answer 31

If your app's entry point js file (i.e. the one you actually run "node" on) is in your project root directory, you can do this really easily with the rootpath npm module. Simply install it via

npm install --save rootpath

...then at the very top of the entry point js file, add:

require('rootpath')();

From that point forward all require calls are now relative to project root - e.g. require('../../../config/debugging/log'); becomes require('config/debugging/log'); (where the config folder is in the project root).

Answer 32

In simple lines, you can call your own folder as module:

For that we need: global and app-module-path module

Here "App-module-path" is the module. It enables you to add additional directories to the Node.js module search path. And "global" is anything that you attach to this object will be available everywhere in your app.

Now take a look at this snippet:

global.appBasePath = __dirname;

require('app-module-path').addPath(appBasePath);

__dirname is the current running directory of Node.js. You can give your own path here to search the path for module.

Answer 33

If you're using ES5 syntax you may use asapp. For ES6 you may use babel-plugin-module-resolver using a config file like this:

.babelrc

{
  "plugins": [
    ["module-resolver", {
      "root": ["./"],
      "alias": {
        "app": "./app",
        "config": "./app/config",
        "schema": "./app/db/schemas",
        "model": "./app/db/models",
        "controller": "./app/http/controllers",
        "middleware": "./app/http/middleware",
        "route": "./app/http/routes",
        "locale": "./app/locales",
        "log": "./app/logs",
        "library": "./app/utilities/libraries",
        "helper": "./app/utilities/helpers",
        "view": "./app/views"
      }
    }]
  ]
}

Answer 34

Some time ago, I created a module for loading modules relative to pre-defined paths.

https://github.com/raaymax/irequire

You can use it instead of 'require'.

irequire.prefix('controllers',join.path(__dirname,'app/master'));
var adminUsersCtrl = irequire("controllers:admin/users");
var net = irequire('net');

Answer 35

Whilst these answers work, they do not address the problem with npm test.

If, for example, I create a global variable in server.js, it will not be set for my test suite execution.

To set a global appRoot variable that will avoid the ../../../ problem and will be available in both npm start and npm test, see:

Mocha tests with extra options or parameters

Note that this is the new official Mocha solution.

Answer 36

Try using asapp:

npm install --save asapp

var { controller, helper, middleware, route, schema, model, APP, ROOT } = require('asapp')

controller('home') instead require('../../controllers/home)

Answer 37

My method of achieving this is to create "local link modules".

Take for example the folder structure of

db ¬
    models ¬
        index.js
    migrations
    seed
    config.json

routes ¬
    index.js
    user ¬
        index.js

If from ./routes/user/index.js I want to access /db/models/index.js, I'd be writing:

require('../../db/models/index.js')

To make the /db/models/index.js accessible from everywhere, I create a folder inside the db folder named _module_, this contains a package.json and a main.js file.

# package.json
{
    "name": "db", <-- change this to what you want your require name to be
    "version": "1.0.0",
    "description": "",
    "author": "",
    "repository": {},
    "main": "main.js"
}

// main.js
module.exports = require('../../db/models/index');

The path in main.js must be relative as if the file is in node_modules, like

node_modules ¬
    db ¬
        main.js

You can then run npm install ./db/_module_ and this will copy the files in ./db/_module_ to ./node_modules/db creating an entry under dependencies in the app's package.json like

"db": "file:db/_module_"

You can now use this package from anywhere by

const db = require('db');

This automatically installs with the rest of your modules when you run npm install, works cross-platform (no symbolic links), and doesn't require any third-party packages.

Answer 38

Since none of the proposed solutions was quite as concise as I wanted, I wanted to share mine here.

I was looking for a solution that doesn't not 'require' anything additional, no modules, no package.json modifications, no environment variables, no external dependencies, minimal CPU cycles, etc.

...so I came up with this:

global.dir = process.cwd();

I do this once when the app loads. Obviously, you can modify the path by adding a subdirectory and/or adding '/..' which will send you up or down the directory tree as needed.

All my includes, in all files, are in this format:

const myFn = require(global.dir + '/js/myDir/myFn');

Answer 39

I don't think you need to solve this in the manner you described. Just use sed if you want to change the same string in a large amount of files. In your example,

find . -name "*.js" -exec sed -i 's/\.\.\/\.\.\//\.\.\//g' {} +

would have ../../ changed to ../

Alternatively, you can require a configuration file that stores a variable containing the path to the library. If you store the following as config.js in the example directory

var config = {};
config.path = '../../';

and in your example file

myConfiguration = require('./config');
express = require(config.path);

You'll be able to control the configuration for every example from one file.

It's really just personal preference.

Answer 40

Many good answers here already. That just shows this is a common problem without clear-cut best solution. Best would be native support in Node.js of course. Here's what I use currently:

const r  = p => require (process.cwd() + p);
let see  = r ('/Subs/SubA/someFile.js' );
let see2 = r ('/Subs/SubB/someFile2.js');
...

I like this solution because the requires-section becomes shorter, no need to type 'require' many times. The main benefit of absolute paths is you can copy them from file to file without having to adjust them like you would with relative paths. Therefore copying also the extra one-liner arrow-function 'r()' is not too much extra work either. And no need to import extra npm-dependencies just to accomplish this very simple task.