Get parent directory name in Node.js

What you want is path.basename:

path.basename(path.dirname(filename))

Daniel Wolf's answer is correct, also if you want the full path of the parent dir:

require('path').resolve(__dirname, '..')

process.mainModule property is deprecated in v14.0.0. If foo.js is run by node foo.js (e.g. somedir/foo.js"),

const path = require("path");

module.exports = path.dirname(require.main.filename);

result: somedir

Use require.main instead

const path = require("path")
path.dirname(path.basename(__dirname))

Whilst the answers herein worked somewhat, I found use of the popular app-root-path module a better anchor point from which to specify a path.

import { path as arp } from 'app-root-path'
import path from 'path'

const root = path.resolve(arp, '../') // the parent of the root path

export const rootDirname = root

Example usage as follows:

import { rootDirname } from './functions/src/utils/root-dirname'
import { getJsonFromFile } from './app/utils/get-json-from-file'

const firebaseJson = getJsonFromFile(`${rootDirname}/firebase.json`)

Maybe not the best answer here but an option not covered by other answers.

The typical __dirname solution doesn't work in an ESM scope. To move one level higher in the directory tree, I came up with the following solution:

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

const __filename = fileURLToPath(import.meta.url);
// First find out the __dirname, then resolve to one higher level in the dir tree
const __dirname = path.resolve(path.dirname(__filename), "../");

If you only need the absolute directory path the file resides in, you can leave out the path.resolve() call.

Using node as of 06-2019, I ran into an issue for accessing just filename. So instead, I just modified it a tiny bit and used:

path.dirname(__filename).split(path.sep).pop()

so now you get the directory name of the current directory you are in and not the full path. Although the previous answers seem to possibly work for others, for me it caused issues as node was looking for a const or a variable but couldn't find one.

const path = require('path');

module.exports = path.dirname(process.mainModule.filename)

Use this anywhere to get the root directory

Simplest way without any node modules like the path. You can easily do in the following manner to get the root folder name.

var rootFolder = __dirname.split('/').pop();
console.log(rootFolder);

__dirname is an environment variable that tells you the absolute path of the directory containing the currently executing file.

Get just the parent with this simple code:

let parentFolder = __dirname.split('\\').pop();

output of my __dirname

console.log({ path: __dirname, parentDirectory: __dirname.split('\\').pop()});

{ 
path: 'C:\\Development\\Guides\\ParentFolder', 
parentDirectory: 'ParentFolder' 
}

Dynamically generates path to parent directory

import {dirname} from "path";
import {fileURLToPath} from "url";
const __dirname = dirname(fileURLToPath(import.meta.url));

If you're import, you can import basename:

import { dirname, basename } from 'node:path';
const parentName = dirname(basename(filename));

use the relative file path from the directory you are working in to access files along with __dirname Here's the documentation