Hướng dẫn nodejs __dirname parent

I am using Node.js, and I want to obtain the parent directory name for a file. I have the file "../test1/folder1/FolderIWant/test.txt".

I want to get "FolderIWant".

I have tried:

var path = require('path');
var parentDir = path.dirname(filename);

But it returns ../test1/folder1/FolderIWant.

asked Mar 22, 2017 at 15:29

What you want is path.basename:

path.basename(path.dirname(filename))

answered May 4, 2017 at 9:57

Daniel WolfDaniel Wolf

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

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

answered Mar 14, 2019 at 15:52

DirigibleDirigible

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

Yeshi

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answered Jul 20, 2020 at 3:03

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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

answered Jul 5, 2020 at 6:23

ByakuByaku

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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.

answered Oct 20, 2021 at 1:53

danday74danday74

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const path = require('path');

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

Use this anywhere to get the root directory

answered Jun 4, 2019 at 18:43

C WilliamsC Williams

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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.

answered Jun 19, 2019 at 8:45

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);

answered Aug 8, 2020 at 11:41

Vinodh RamVinodh Ram

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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.

answered Aug 1 at 10:40

Tom BöttgerTom Böttger

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