Question 224252 in Javascript Question posted in
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Javascript – How to export for specific files in nodejs or deno

By using export keyword, we can export anything from a file to another file in JavaScript. But is their any way to export for specific files. So that other file can’t import from this file.

export const t = {
a:'this will export for index.js only'
}
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2

Answers


  1. 0

    If you are referring to how java controls access of classes to the rest of the application public, private, protected then no. If you only want a singular file to be able to use a variable then declare the variable in the file you want to use. or look into changing the architecture of your application. Ex: change your function to use dependency injection.

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  2. 0

    There’s no way to achieve that directly, since import.meta does not contain that information. But you can do it if you export a factory function instead, that way you can access the caller file by using Error.stack.

    module.js

    const callerUrl = () => {
  const error = new Error()
  const lines = error.stack?.split('n')
  let line = lines && lines[3] || ''
  line = line.replace('at', '')
  line = line.replace(/:[0-9]+:[0-9]+/, '')
  return line.trim()
}

export default () => {
  const file = callerUrl();
  const x = {};

  if(file.endsWith('/index.js')) {
    x.a = 'from index.js'
  }
  else if(file.endsWith('/foo.js')) {
    x.a = 'from foo.js'
  }

  return x;
}

    index.js

    import factory from './module.js';
import './foo.js';

console.log(factory());
// { a: "from index.js" }

    See: Access the calling file name from within a function in a different file

    I don’t recommend exporting something based on the caller file though.

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