node.js equivalent of python's if name == 'main'

Question

I'd like to check if my module is being included or run directly. How can I do this in node.js?

@DimaTisnek Source or any more info on `isMain`? It sounds fantastic but I can't find anything about it — Ken Bellows, Oct 02 '20 at 12:00
The only significant reference I can find is in [a Gist from CJ Silveiro](https://gist.github.com/ceejbot/b49f8789b2ab6b09548ccb72813a1054) describing NPM's proposal/vision for ESM modules in Node. I haven't been able to find anything official from Node.js themselves. Any links would be appreciated — Ken Bellows, Oct 02 '20 at 12:15
https://exploringjs.com/nodejs-shell-scripting/ch_nodejs-path.html#detecting-if-module-is-main detecting if the current module is “main” (the app entry point) — aderchox, Apr 20 '23 at 10:51

score 467 · Accepted Answer · edited May 02 '23 at 05:37

467

The nodejs docs describe another way to do this which may be the preferred method:

When a file is run directly from Node, require.main is set to its module.

To take advantage of this, check if this module is the main module and, if so, call your main code:

function myMain() {
    // main code
}

if (require.main === module) {
    myMain();
}

EDIT: If you use this code in a browser, you will get a "Reference error" since "require" is not defined. To prevent this, use:

if (typeof require !== 'undefined' && require.main === module) {
    myMain();
}

edited May 02 '23 at 05:37

Shlomo

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answered May 22 '11 at 19:43

Stephen Emslie

10,539
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18

you always have to check require.main === module **irrespective** of your function name. To make it clear above code should be modified as: `var fnName = function(){ // code } if (require.main === module) { fnName(); }` – Kunal Kapadia May 29 '15 at 12:47
I would wrap it with try...catch for browser compatibility – Ohad Cohen May 23 '16 at 14:14
4

@OhadCohen "try...catch" might also catch real errors. I think it is better to just check if typeof require != 'undefined'. – Erel Segal-Halevi Apr 27 '17 at 19:17
But what if the file is forked? – deostroll May 03 '17 at 03:11
4

Sadly this feature is gone with `--experimental-modules`. – seamlik Jul 19 '18 at 14:06
2

@seamlik in this case, the package [es-main](https://www.npmjs.com/package/es-main) can be used. Usage: `esMain(import.meta)` returns true iff the file is not imported. – Charlie Harding May 13 '21 at 12:05
For the deno users that came here https://deno.land/manual@v1.0.0/examples/testing_if_main – Norfeldt Oct 22 '21 at 10:08

nornagon · Answer 2 · 2019-08-13T20:36:16.517

69

if (!module.parent) {
  // this is the main module
} else {
  // we were require()d from somewhere else
}

EDIT: If you use this code in a browser, you will get a "Reference error" since "module" is not defined. To prevent this, use:

if (typeof module !== 'undefined' && !module.parent) {
  // this is the main module
} else {
  // we were require()d from somewhere else or from a browser
}

edited Aug 13 '19 at 20:36

answered Feb 13 '11 at 01:54

nornagon

15,393
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Is this documented somewhere? – intuited May 18 '11 at 03:14
11

Nope, but it's used in one of [node.js's tests](https://github.com/joyent/node/blob/master/test/simple/test-cli-eval.js) – nornagon May 19 '11 at 08:52
1

To me this reads better than the accepted answer and has the benefit of not requiring the module's "name" – blented Sep 20 '13 at 18:47
12

the accepted answer doesn't use the module's name either. – nornagon Sep 20 '13 at 22:53
2

this is deprecated. use `module.main` now – pcnate Sep 15 '21 at 21:18
1

@pcnate Could you consider adding another answer to explain how to use `module.main`? – tsh Nov 11 '21 at 02:24

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