Node.js - getting current filename

Question

How to get current file name, function name and line number?

I want to use it for logging/debugging purpose, equivalent to __FILE__, __LINE__ in c

File: http://stackoverflow.com/questions/3133243/how-to-get-path-to-current-script-with-node-js?rq=1 Line: https://github.com/visionmedia/callsite — Matt Ball, Jan 07 '13 at 18:04
I see nobody ever came up with a good answer for function name or line number. — Mr. Lance E Sloan, Apr 27 '21 at 20:27

herve · Answer 1 · 2015-03-06T16:27:10.587

305

Node.js provides a standard API to do so: Path.

Getting the name of the current script is then easy:

var path = require('path');
var scriptName = path.basename(__filename);

edited Mar 06 '15 at 16:27

answered Feb 04 '15 at 10:31

herve

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Doesn't this get the basename rather than the actual filename? – Phil May 11 '17 at 07:30
16

for a script named `script.js`, you'll get `script.js`. If you want to have the name only without extension, as we are speaking here of a `js` script you can use `var scriptName = path.basename(__filename, '.js');` or `var scriptName = path.basename(__filename, path.extname(__filename))` – herve May 12 '17 at 08:15
11

another way to drop the extension: `path.parse(__filename).name` – RH Becker Feb 05 '18 at 20:59
11

`__filename` is not defined in an ES6 Module. – Cullub Jun 24 '20 at 19:15
6

@Cullub, extract from the nodejs [documentation](https://nodejs.org/api/esm.html#esm_no_require_exports_module_exports_filename_dirname): These `CommonJS` variables are not available in ES modules. Equivalents of `__filename` and `__dirname` can be created inside of each file via `import.meta.url`. – herve Jun 27 '20 at 12:50
You can remove the extension by regex string replace too. and it can handle upper case letter in filename `var scriptName = path.basename(__filename).replace(/\.js$/i, '' ); ` – fedmich Sep 03 '21 at 02:45
@Cullub `module.filename` seems to work for me. – Asaf Jul 27 '22 at 16:15

Anthony McCormick · Answer 2 · 2014-03-30T21:32:17.717

58

within a module you can do any of the following to get the full path with filename

this.filename;
module.filename;
__filename;

If you just want the actual name with no path or extension you can do something like this.

module.filename.slice(__filename.lastIndexOf(path.sep)+1, module.filename.length -3);

edited Mar 30 '14 at 21:32

answered Sep 22 '13 at 10:02

Anthony McCormick

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4

You probably want to use path.sep to make it cross platform. Here's what I did __filename.slice(__filename.lastIndexOf(path.sep)+1); – Ryan Mar 29 '14 at 21:28
2

For those who just want to copy and paste this super handy one-liner, don't forget to either `var path = require('path');` beforehand or change it to `module.filename.slice(__filename.lastIndexOf(require('path').sep)+1, module.filename.length -3);` – forrestranger Aug 19 '14 at 00:23
22

the -3 assumes a length of extension, something like this might work better `path.basename(module.filename, path.extname(module.filename))` – dre Sep 02 '14 at 21:01

Red Walrus · Answer 3 · 2015-08-06T12:51:06.467

52

To get the file name only. No additional module simply:

// abc.js
console.log(__filename.slice(__dirname.length + 1));

 // abc
console.log(__filename.slice(__dirname.length + 1, -3));

edited Aug 06 '15 at 12:51

answered Aug 06 '15 at 12:45

Red Walrus

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1

This is what I came here for. Thanks! – leviathanbadger Mar 12 '17 at 00:24
1

This should be the accepted answer. Works in strict mode. Very straightforward. Thank you! – Matthew Mar 22 '17 at 22:22
7

Unless your extension isn't 2 characters long, which is totally possible because you can bind script interpretation to any file extension you want... so this may not be a good all-around solution. – ErikE Mar 14 '19 at 01:57
I literally copied and pasted this. Thank you – Alex Antoine Sep 22 '22 at 20:36

score 26 · Answer 4 · edited May 23 '17 at 12:18

26

'use strict';

const scriptName = __filename.split(/[\\/]/).pop();

Explanation

console.log(__filename);
// 'F:\__Storage__\Workspace\StackOverflow\yourScript.js'
const parts = __filename.split(/[\\/]/);
console.log(parts);
/*
 * [ 'F:',
 *   '__Storage__',
 *   'Workspace',
 *   'StackOverflow',
 *   'yourScript.js' ]
 *
**/

Here we use split function with regular expression as the first parameter.
The regular expression we want for separator is [\/] (split by / or \) but / symbol must be escaped to distinguish it from regex terminator /, so /[\\/]/.

const scriptName = __filename.split(/[\\/]/).pop(); // Remove the last array element
console.log(scriptName);
// 'yourScript.js'

Don't Use This

You really should use path.basename instead (first documented in Node.js v0.1.25), because it handles all the corner cases you don't want to know about like filenames with slashes inside (e.g. file named "foo\bar" on unix). See path answer above.

edited May 23 '17 at 12:18

Community

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answered Aug 31 '15 at 11:41

ilyaigpetrov

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Please add explanation – Daniel Cheung Aug 31 '15 at 12:40
1

@DanielCheung Here you are! But really all this is searchable on the Internets, just google `javascript split` and get going. – ilyaigpetrov Sep 01 '15 at 13:28
8

I was helping you... -_- Your answer was being voted for deletion. An answer not explaining enough or just containing code **will** get deleted. – Daniel Cheung Sep 01 '15 at 14:33

score 3 · Answer 5 · answered Jan 08 '13 at 07:26

3

You might also look at console-plus. This adds filename and linenumber to any logging text and has different colours for .log, .info and .error.

answered Jan 08 '13 at 07:26

heinob

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Pretty cool. Install with "npm install --save console-plus", or if you're using yarn: "yarn add console-plus" – Andrew Faulkner Dec 22 '16 at 18:53

score 1 · Answer 6 · answered Apr 03 '17 at 18:58

1

Well, I wanted to use the file name as the tag for the logger so the most straight forward and what I wanted is this:

__filename.substring(__dirname.length + 1, __filename.lastIndexOf("."))

answered Apr 03 '17 at 18:58

TacB0sS

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score 1 · Answer 7 · answered Jul 10 '18 at 10:20

1

I know it's been already a long time, but what I did is __filename.split('\\').pop(). This will get the full path with the filename, split it by \ then get the last index which will be your filename.

answered Jul 10 '18 at 10:20

Marcus

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On macOS `__filename.split('/').pop()` – Reece Daniels Jun 07 '19 at 22:21
4

On any system `require('path').basename(__filename)` – Cameron Tacklind Jul 07 '21 at 02:45

score 1 · Answer 8 · answered Jan 27 '19 at 15:32

you can use this function to get filename:

/**
 * @description 
 * get file name from path
 * @param {string} path path to get file name
 * @returns {string} file name
 * @example 
 * getFileName('getFileName/index.js') // => index.js
 */
export default function getFileName(path) {
    return path.replace(/^.*[\\\/]/, '');
}

if you use nodejs, you can install the package that include this function https://www.npmjs.com/package/jotils

score 1 · Answer 9 · answered Jan 09 '22 at 19:06

Only because I don't see it listed here, another option is to just remove the __dirname from the __filename.

__filename.replace(`${__dirname}/`, '')

I think that is useful if you don't want to import a the path module. However, I do believe that path.basename(__filename) is the most appropriate.

Node.js - getting current filename

9 Answers9

Explanation

Don't Use This

Linked