`if __name__ == '__main__'` equivalent in javascript es6 modules

Solution 1:

An alternative for ES6 modules is now supported in Node. Using the new import.meta builtin. (Don't forget to set "type": "module" in package.json.)

Example

// main.js
import "./lib.js"
import { fileURLToPath } from "url";

if (process.argv[1] === fileURLToPath(import.meta.url)) {
  console.log("I print to stdout!");
}

// lib.js
import { fileURLToPath } from "url";

if (process.argv[1] === fileURLToPath(import.meta.url)) {
  console.log("I don't run, because I'm an imported module.");
}

$ node main.js output:

I print to stdout!

Utility function

I like to just import { isMain } from "./lib/utils.js" and pass import.meta.url to isMain().

import { argv } from "process"
import { fileURLToPath } from "url"
/**
 * Check if a module is the main module launched with the node process.
 * Meaning the module is NOT imported by another module,
 * but was directly invoked by `node`, like this: `$ node main.js`
 *
 * @example
 * ```js
 * // main.js
 * import lib from "./lib.js"
 * import { isMain } from "./utils.js"
 *
 * if (isMain(import.meta.url)) {
 *   console.log("I print to stdout")
 * }
 *
 * // lib.js
 * import { isMain } from "./utils"
 *
 * if (isMain(import.meta.url)) {
 *   console.log("I don't run, because I'm an imported module")
 * }
 * ```
 *
 * @param {string} moduleUrl needs to be `import.meta.url`
 * @returns {boolean} true if the module is the main module
 */
export function isMain(moduleUrl) {
  const modulePath = fileURLToPath(moduleUrl)
  const [_binPath, mainScriptPath] = argv
  return modulePath === mainScriptPath
}

Solution 2:

module.parent will help you:

if(module.parent) {
    console.log('required module')
} else {
    console.log('main')
}