Alternative for __dirname in Node.js when using ES6 modules

Question

I use the flag --experimental-modules when running my Node application in order to use ES6 modules.

However when I use this flag the metavariable __dirname is not available. Is there an alternative way to get the same string that is stored in __dirname that is compatible with this mode?

Answer 1

As of Node.js 10.12 there's an alternative that doesn't require creating multiple files and handles special characters in filenames across platforms:

import { dirname } from 'path';
import { fileURLToPath } from 'url';

const __dirname = dirname(fileURLToPath(import.meta.url));

Answer 2

The most standardized way in 2021

Note: it doesn't work in Windows.

import { URL } from 'url'; // in Browser, the URL in native accessible on window

const __filename = new URL('', import.meta.url).pathname;
// Will contain trailing slash
const __dirname = new URL('.', import.meta.url).pathname;

And forget about join to create paths from the current file, just use the URL

const pathToAdjacentFooFile = new URL('./foo.txt', import.meta.url).pathname;
const pathToUpperBarFile = new URL('../bar.json', import.meta.url).pathname;

Answer 3

For Node 10.12 +...

Assuming you are working from a module, this solution should work, and also gives you __filename support as well

import path from 'node:path';
import { fileURLToPath } from 'node:url';

const __filename = fileURLToPath(import.meta.url);
const __dirname = path.dirname(__filename);

The nice thing is that you are also only two lines of code away from supporting require() for CommonJS modules. For that you would add:

import { createRequireFromPath } from 'module';
const require = createRequireFromPath(__filename); 

Answer 4

In most cases, using what is native to Node.js (with ES Modules), not external resources, the use of __filename and __dirname for most cases can be totally unnecessary. Most (if not all) of the native methods for reading (streaming) supports the new URL + import.meta.url, exactly as the official documentation itself suggests:

• No __filename or __dirname
• No JSON Module Loading
• No require.resolve

As you can see in the description of the methods, the path parameter shows the supported formats, and in them include the <URL>, examples:

Method	path param supports
fs.readFile(path[, options], callback)	<string>, <Buffer>, <URL>, <integer>
fs.readFileSync(path[, options])	<string>, <Buffer>, <URL>, <integer>
fs.readdir(path[, options], callback)	<string>, <Buffer>, <URL>
fs.readdirSync(path[, options])	<string>, <Buffer>, <URL>, <integer>
fsPromises.readdir(path[, options])	<string>, <Buffer>, <URL>
fsPromises.readFile(path[, options])	<string>, <Buffer>, <URL>, <FileHandle>

So with new URL('<path or file>', import.meta.url) it solves and you don't need to be treating strings and creating variables to be concatenated later.

Examples:

See how it is possible to read a file at the same level as the script without needing __filename or any workaround:

import { readFileSync } from 'fs';

const output = readFileSync(new URL('./foo.txt', import.meta.url));

console.log(output.toString());

List all files in the script directory:

import { readdirSync } from 'fs';

readdirSync(new URL('./', import.meta.url)).forEach((dirContent) => {
  console.log(dirContent);
});

Note: In the examples I used the synchronous functions just to make it easier to copy and execute.

If the intention is to make a "own log" (or something similar) that will depend on third parties, it is worth some things done manually, but within the language and Node.js this is not necessary, with ESMODULES it is totally possible not to depend on either __filename and neither __dirname, since native resources with new URL with already solve it.

---

Note that if you are interested in using something like require at strategic times and need the absolute path from the main script, you can use module.createRequire(filename) (Node.js v12.2.0 + only) combined with import.meta.url to load scripts at levels other than the current script level, as this already helps to avoid the need for __dirname, an example using import.meta.url with module.createRequire:

import { createRequire } from 'module';

const require = createRequire(import.meta.url);

// foo-bar.js is a CommonJS module.
const fooBar = require('./foo-bar');

fooBar();

Source from foo-bar.js:

module.exports = () => {
    console.log('hello world!');
};

Which is similar to using without "ECMAScript modules":

const fooBar = require('./foo-bar');

Answer 5

There have been proposals about exposing these variables through import.meta, but for now, you need a hacky workaround that I found here:

// expose.js
module.exports = {__dirname};

// use.mjs
import expose from './expose.js';
const {__dirname} = expose;

Answer 6

I used:

import path from 'path';

const __dirname = path.resolve(path.dirname(decodeURI(new URL(import.meta.url).pathname)));

decodeURI was important: used spaces and other stuff within the path on my test system.

path.resolve() handles relative urls.

edit:

fix to support windows (/C:/... => C:/...):

import path from 'path';

const __dirname = (() => {let x = path.dirname(decodeURI(new URL(import.meta.url).pathname)); return path.resolve( (process.platform == "win32") ? x.substr(1) : x ); })();

Answer 7

I made this module es-dirname that will return the current script dirname.

import dirname from 'es-dirname'

console.log(dirname())

It works both in CommonJs scripts and in ES Modules both on Windows and Linux.

Open an issue there if have an error as the script has been working so far in my projects but it might fail in some other cases. For this reason do not use it in a production environment. And this is a temporary solution as I am sure the Node.js team will release a robust way to do it in a near future.

Answer 8

import path from 'path';
import { fileURLToPath } from 'url';

const __filename = fileURLToPath(import.meta.url);
const __dirname = path.dirname(__filename);


// do not use the following code which is bad for CJK characters
const __filename = new URL('', import.meta.url).pathname;

Answer 9

import path from 'path';
const __dirname = path.join(path.dirname(decodeURI(new URL(import.meta.url).pathname))).replace(/^\\([A-Z]:\\)/, "$1");

This code also works on Windows. (the replacement is safe on other platforms, since path.join returns back-slash separators only on Windows)

Answer 10

Since other answers, while useful, don't cover both cross-platform cases (Windows POSIX) and/or path resolution other than the __dirname or __filename and it's kind of verbose to repeat this kind of code everywhere:

import { dirname, join } from 'path'
import { fileURLToPath } from 'url'

const __filename = fileURLToPath(import.meta.url)
const __dirname = dirname(__filename)

const somePath = join(__dirname, '../some-dir-or-some-file')

I just published a NPM package called esm-path to help with this kind of recurring task, hoping it can also be useful to others.

It's documented but here how to use it:

import { getAbsolutePath } from 'esm-path'

const currentDirectoryPath = getAbsolutePath(import.meta.url)
console.log(currentDirectoryPath)

const parentDirectoryPath = getAbsolutePath(import.meta.url, '..')
console.log(parentDirectoryPath)

// Adapt the relative path to your case
const packageJsonFilePath = getAbsolutePath(import.meta.url, '../package.json')
console.log(packageJsonFilePath)

// Adapt the relative path to your case
const packageJsonFilePath = getAbsolutePath(import.meta.url, '..' , 'package.json')
console.log(packageJsonFilePath)

Answer 11

Just use path.resolve() method.

import { resolve } from 'path';

app.use('/public/uploads', express.static(resolve('public', 'uploads')))

Answer 12

I use this option, since the path starts with file:// just remove that part.

const __filename = import.meta.url.slice(7);
const __dirname = import.meta.url.slice(7, import.meta.url.lastIndexOf("/"));

Answer 13

As Geoff pointed out the following code returns not the module's path but working directory.

import path from 'path';
const __dirname = path.resolve();

works with --experimental-modules

Answer 14

create a file called root-dirname.js in your project root with this in it:

import { dirname } from 'path'

const dn = dirname(new URL(import.meta.url).hostname)
const __dirname = process.platform === 'win32' ? dn.substr(1) : dn // remove the leading slash on Windows
export const rootDirname = __dirname

Then just import rootDirname when you want the path to the project root folder.

Other than that, Rudolf Gröhling's answer is also correct.

Answer 15

Agree or disagree with the use of global, I found this to be the easiest way to remember and refactor existing code.

Put somewhere early in your code execution:

import { fileURLToPath } from 'node:url';
import { dirname } from 'node:path';

global.___filename = (path) => {
  return fileURLToPath(path);
};

global.___dirname = (path) => {
  return dirname(global.___filename(path));
};

And then in whichever file you need dirname or filename:

___filename(import.meta.url)
___dirname(import.meta.url)

Of course if we had macros, I wouldn't need to pass import.meta.url, perhaps there's an improvement.

Answer 16

This works:

import path from 'node:path';
import url from 'node:url';

const DIR_NAME = path.dirname(url.fileURLToPath(import.meta.url));

Answer 17

This is how I solved it for quicker access.

import { dirname } from 'path'
import { fileURLToPath } from 'url'

/**
 * ES6 __filename polyfill
 *
 * @param {String} fileLocation - Use `import.meta.url`
 * @returns {String} - File path
 * @example
 * const __filename = _filename(import.meta.url)
 */
export const _filename = fileLocation => {
  return fileURLToPath(fileLocation)
}

/**
 * ES6 __dirname polyfill
 *
 * @param {String} fileLocation - Use `import.meta.url`
 * @returns {String} - Directory path
 * @example
 * const __dirname = _dirname(import.meta.url)
 */
export const _dirname = fileLocation => {
  return dirname(fileURLToPath(fileLocation))
}

Save this as es6-polyfills.js and then import _filename and _dirname where required. Remember to set __dirname and __filename as described in comments.

Answer 18

You can use the common-es package exactly for that:

npm i common-es

Usage

// # myProjectFile.js or myProjectfile.ts

import { getGlobals } from 'common-es'
const { __dirname, __filename } = getGlobals(import.meta.url)
// now you can use __dirname or file name normally as you would do in commonjs

If you're not familiar with this weird syntax, read this reference on MDN Here

Answer 19

another option

import {createRequire} from 'module'; // need node v12.2.0

const require = createRequire(import.meta.url);
const __dirname = require.resolve.paths('.')[0];

Answer 20

You can use the stack from a new Error(). The error doesn't need to be thrown, and won't stop program execution either. The first line of the stack will always be the error and its message, with the second line being the file the which the error was invoked from.

Since this is a method (which is probably in a util.js file), the real location of the getDirname() call is actually the third line of the error stack.

export const getDirname = () => {
    // get the stack
    const { stack } = new Error();
    // get the third line (the original invoker)
    const invokeFileLine = stack.split(`\n`)[2];
    // match the file URL from file://(.+)/ and get the first capturing group
    //     the (.+) is a greedy quantifier and will make the RegExp expand to the largest match
    const __dirname = invokeFileLine.match(/file:\/\/(.+)\//)[1];
    return __dirname;
};

Answer 21

I have also published a package on NPM called cross-dirname (forked from es-dirname). The package is tested with Node.js (ESM and CJS), Deno and GJS.

Example:

import dirname from 'cross-dirname'

console.log(dirname())

Answer 22

i made 1 file const.js constaining :

Object.assign(global, {
    __dirname: __dirname,
    __filename: __filename,
});

then

import './const.js';
console.log(__dirname);

Answer 23

Get catch-all issue has been resolved and worked for me:

Syntax: In my server.js

const express = require("express");

const path = require('path');

__dirname = path.resolve();

const app = express();

app.get('/*', function(req, res) { res.sendFile(path.join(__dirname, './public/index.html'), function(err) { if (err) { res.status(500).send(err) } }) })

Answer 24

process.cwd()

From documentation:

The process.cwd() method returns the current working directory of the Node.js process.