Detect if called through require or directly by command line

For those using ES Modules (and Node 10.12+), you can use import.meta.url:

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

const nodePath = path.resolve(process.argv[1]);
const modulePath = path.resolve(fileURLToPath(import.meta.url))
const isRunningDirectlyViaCLI = nodePath === modulePath

Things like require.main, module.parent and __dirname/__filename aren’t available in ESM.

Note: If using ESLint it may choke on this syntax, in which case you’ll need to update to ESLint ^7.2.0 and turn your ecmaVersion up to 11 (2020).

More info: process.argv, import.meta.url

I always find myself trying to recall how to write this goddamn code snippet, so I decided to create a simple module for it. It took me a bit to make it work since accessing caller's module info is not straightforward, but it was fun to see how it could be done.

So the idea is to call a module and ask it if the caller module is the main one. We have to figure out the module of the caller function. My first approach was a variation of the accepted answer:

module.exports = function () {
    return require.main === module.parent;
};

But that is not guaranteed to work. module.parent points to the module which loaded us into memory, not the one calling us. If it is the caller module that loaded this helper module into memory, we're good. But if it isn't, it won't work. So we need to try something else. My solution was to generate a stack trace and get the caller's module name from there:

module.exports = function () {
    // generate a stack trace
    const stack = (new Error()).stack;
    // the third line refers to our caller
    const stackLine = stack.split("\n")[2];
    // extract the module name from that line
    const callerModuleName = /\((.*):\d+:\d+\)$/.exec(stackLine)[1];

    return require.main.filename === callerModuleName;
};

Save this as is-main-module.js and now you can do:

const isMainModule = require("./is-main-module");

if (isMainModule()) {
    console.info("called directly");
} else {
    console.info("required as a module");
}

Which is easier to remember.

Try this if you are using ES6 modules:

if (process.mainModule.filename === __filename) {
  console.log('running as main module')
}

First, let's define the problem better. My assumption is that what you are really looking for is whether your script owns process.argv (i.e. whether your script is responsible for processing process.argv). With this assumption in mind, the code and tests below are accurate.

module.parent works excellently, but it is deprecated for good reasons (a module might have multiple parents, in which case module.parent only represents the first parent), so use the following future-proof condition to cover all cases:

if (
  typeof process === 'object' && process && process.argv
   && (
    (
      typeof module === 'object' && module
       && (
        !module.parent
         || require.main === module
         || (process.mainModule && process.mainModule.filename === __filename)
         || (__filename === "[stdin]" && __dirname === ".")
       )
    )
    || (
      typeof document === "object"
      && (function() {
       var scripts = document.getElementsByTagName("script");
       try { // in case we are in a special environment without path
         var normalize = require("path").normalize;
         for (var i=0,len=scripts.length|0; i < len; i=i+1|0)
           if (normalize(scripts[i].src.replace(/^file:/i,"")) === __filename)
             return true;
       } catch(e) {}
      })()
    )
   )
) {
    // this module is top-level and invoked directly by the CLI
    console.log("Invoked from CLI");
} else {
    console.log("Not invoked from CLI");
}

It works correctly in all of the scripts in all of the following cases and never throws any errors^†:

• Requiring the script (e.x. require('./main.js'))
• Directly invoking the script (e.x. nodejs cli.js)
• Preloading another script (e.x. nodejs -r main.js cli.js)
• Piping into node CLI (e.x. cat cli.js | nodejs)
• Piping with preloading (e.x. cat cli.js | nodejs -r main.js)
• In workers (e.x. new Worker('./worker.js'))
• In evaled workers (e.x. new Worker('if (<test for CLI>) ...', {eval: true}))
• Inside ES6 modules (e.x. nodejs --experimental-modules cli-es6.js)
• Modules with preload (e.x. nodejs --experimental-modules -r main-es6.js cli-es6.js)
• Piped ES6 modules (e.x. cat cli-es6.js | nodejs --experimental-modules)
• Pipe+preload module (e.x. cat cli-es6.js | nodejs --experimental-modules -r main-es6.js)
• In the browser (in which case, CLI is false because there is no process.argv)
• In mixed browser+server environments (e.x. ElectronJS, in which case both inline scripts and all modules loaded via <script> tags are considered CLI)

The only case where is does not work is when you preload the top-level script (e.x. nodejs -r cli.js cli.js). This problem cannot be solved by piping (e.x. cat cli.js | nodejs -r cli.js) because that executes the script twice (once as a required module and once as top-level). I do not believe there is any possible fix for this because there is no way to know what the main script will be from inside a preloaded script.

^† Theoretically, errors might be thrown from inside of a getter for an object (e.x. if someone were crazy enough to do Object.defineProperty(globalThis, "process", { get(){throw 0} });), however this will never happen under default circumstances for the properties used in the code snippet in any environment.

How can I detect whether my node.js file was called directly from console (windows and unix systems) or loaded using the ESM module import ( import {foo} from 'bar.js')

Such functionality is not exposed. For the moment you should separate your cli and library logic into separate files.

Answer from node.js core contributor devsnek replying to nodejs/help/issues/2420

It's the right answer in my point of view